User:Sudoku's Integral

1.$$\int_{0}^{1}\int_{0}^{1}\frac{1-(xy)^n}{1-xy}\mathrm dx \mathrm dy=\color{red}H_n^2$$

2.$$\int_{0}^{1}\int_{0}^{1}\frac{1-(xy)^n}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{blue}H_n$$

3.$$\int_{0}^{1}\int_{0}^{1}\frac{1-(xy)^n}{1-xy}\cdot \frac{\ln y}{\ln^2(xy)}\mathrm dx \mathrm dy=-\color{green}\frac{1}{2}H_n$$

4.$$\int_{0}^{1}\int_{0}^{1}\frac{1-(xy)^n}{1-xy}\cdot \frac{x}{\ln(xy)}\mathrm dx \mathrm dy=-\color{purple}\ln(n+1)$$

5.$$\int_{0}^{1}\int_{0}^{1}\frac{1-(xy)^v}{1-xy}\cdot \frac{x^n}{\ln(xy)}\mathrm dx \mathrm dy=-\color{purple}\frac{1}{n}\ln\left(\frac{\color{red}\Gamma(v+n+1)}{v\Gamma(v)\Gamma(n+1)}\right)$$

6.$$\int_{0}^{1}\int_{0}^{1}\frac{1-(xy)^n}{(1-xy)^2}\cdot \frac{\mathrm dx \mathrm dy}{x^{-1}}=\color{red}H_n$$

7.$$\int_{0}^{1}\int_{0}^{1}\frac{1-xy}{1+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=\color{red}\ln\left(\color{green}\frac{\pi}{2}\right)$$

8.$$\int_{0}^{1}\int_{0}^{1}\frac{1-xy}{1+xy}\cdot \frac{xy}{\ln^2(xy)}\mathrm dx \mathrm dy=\color{red}\ln\left(\color{blue}\frac{4}{\pi}\right)$$

9.$$\int_{0}^{1}\int_{0}^{1}\frac{1-xy}{1+xy}\cdot \frac{\ln x}{\ln^2(xy)}\mathrm dx \mathrm dy=\frac{1}{2}-\color{red}\ln 2$$

10.$$\int_{0}^{1}\int_{0}^{1}\frac{1-xy}{1+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=1-2\color{red}\ln 2$$

11.$$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2y^2}{1-x^2y^2}\cdot \left(\frac{y}{x}\right)^{2\alpha}\mathrm dx \mathrm dy=\color{red}\frac{1}{4\alpha^2-1}$$, $$\alpha\ge 1$$

12.$$\int_{0}^{1}\int_{0}^{1}\frac{1-x^2y^2}{1+x^2y^2}\cdot \frac{xy}{\ln(xy)}\mathrm dx \mathrm dy=\color{red}\frac{1}{2}-\ln 2$$

13.$$\int_{0}^{1}\int_{0}^{1}\frac{1-x^2y^2}{1+x^2y^2}\cdot \frac{xy}{\ln^2(xy)}\mathrm dx \mathrm dy=\color{green}\ln\left(\frac{\pi}{2}\right)$$

14.$$\int_{0}^{1}\int_{0}^{1}\frac{1-x^2y^2}{1+x^2y^2}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{purple}\frac{2-\pi}{2}$$

15.$$\int_{0}^{1}\int_{0}^{1}\frac{1-x^2y^2}{1+x^2y^2}\cdot \frac{\ln x}{\ln(xy)}\mathrm dx \mathrm dy=\color{blue}\frac{2C-1}{2}$$

16.$$\int_{0}^{1}\int_{0}^{1}\frac{1-x^2y^2}{1+x^2y^2}\cdot \frac{\ln x}{\ln^2(xy)}\mathrm dx \mathrm dy=\color{brown}\frac{2-\pi}{4}$$

17.$$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2y^2}{1-x^2y^2}\cdot \frac{\ln x}{\ln(xy)}\mathrm dx \mathrm dy=\color{blue}\frac{\pi^2}{8}-\frac{1}{2}$$

18.$$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2y^2}{(1-xy)^2}\cdot \frac{y}{x} \cdot \ln x \mathrm dx \mathrm dy=-\color{blue}\frac{3}{2}$$

19.$$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2y^2}{(1-xy)^2}\cdot \ln x \mathrm dx \mathrm dy=\color{blue}2\zeta(3)-2\zeta(2)+2\zeta(0)$$

20.$$\int_{0}^{1}\int_{0}^{1}\frac{(1-xy)^2}{1+x^2y^2}\cdot \frac{y}{x} \cdot \ln x \mathrm dx \mathrm dy=\color{blue}\beta(2)-\color{red}\beta(1)+\color{green}\beta(0)$$

21.$$\int_{0}^{1}\int_{0}^{1}\frac{(1-xy)^2}{1+x^2y^2}\cdot \frac{y}{x} \cdot \ln y \mathrm dx \mathrm dy=\color{blue}\beta(2)+\color{red}\beta(1)-\color{green}3\beta(0)$$

22.$$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2y^2}{(1-xy)^2}\cdot \ln^{2n-1}(x) \mathrm dx \mathrm dy=\color{red}2\Gamma(2n)[\color{blue}\zeta(2n+1)-\zeta(2n)+\zeta(0)]$$

23.$$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2y^2}{(1-xy)^2}\cdot \ln^{2n}(x) \mathrm dx \mathrm dy=\color{red}2\Gamma(2n+1)[\color{blue}\zeta(2n+1)-\zeta(2n+2)-\zeta(0)]$$

24.$$\int_{0}^{1}\int_{0}^{1}\frac{(1-xy)^2}{1+x^2y^2}\cdot \frac{y}{x} \cdot \ln(xy) \mathrm dx \mathrm dy=\color{blue}2C-1$$

25.$$\int_{0}^{1}\int_{0}^{1}\frac{(1-xy)^2}{1+x^2y^2}\cdot \frac{y}{x} \cdot \ln\left(\frac{y}{x}\right) \mathrm dx \mathrm dy=\color{blue}2\beta(1)-2$$

26.$$\int_{0}^{1}\int_{0}^{1}\frac{(1-x)(\alpha+\beta x)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{red}\left(\color{blue}\alpha \gamma+\frac{\beta}{2}\ln 2\right)$$

27.$$\int_{0}^{1}\int_{0}^{1}\frac{(1-x)(\alpha-\beta xy)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{red}\left[\color{blue}(\alpha-\beta)\gamma+\beta(1-\ln 2)\right]$$

28.$$\int_{0}^{\infty}\frac{x-1}{\sqrt{2^x+1}}\cdot \frac{\mathrm dx}{\ln(2^x+1)}=\color{blue}\frac{4}{\pi^2+4}\cdot \color{red}\frac{1}{\ln^2(2)}$$

29.$$\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\cdot \ln\left(\frac{1+x}{1-x}\right)\ln\left(\frac{\alpha+x}{\alpha-x}\right)\mathrm dx=\color{red}{\pi}\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{\alpha+x}{\alpha-x}}\cdot \ln\left(\frac{\alpha+x}{\alpha-x}\right)$$

30.$$\int_{0}^{\pi}\sin^2(x)\ln^{n}\left[\cos\left(\frac{x}{2}\right)\right]\frac{\mathrm dx}{\sin\left(\frac{x}{2}\right)}=(-1)^n\color{blue}\frac{2^3}{3}\cdot \color{red}\frac{n!}{3^n}$$

31.$$\int_{0}^{\pi}\sin(x)\ln^{n}\left[\sin\left(\frac{x}{2}\right)\right]\frac{\mathrm dx}{\sin\left(\frac{x}{2}\right)}=4(-1)^n\color{red} n!$$

32.$$\int_{\pi/4}^{\pi/2}\sin(2x)\ln\ln\tan x\mathrm dx=-\frac{1}{2}\color{red}\left(\color{blue}\gamma+\ln\frac{4}{\pi}\right)=-\frac{\gamma}{2}+\ln\Gamma(3/2)$$

33.$$\int_{0}^{\pi/4}\sin(2x)\ln\ln\cot x\mathrm dx=-\frac{1}{2}\color{brown}\left(\color{blue}\gamma+\ln\frac{4}{\pi}\right)$$

34.$$\int_{0}^{\pi/4}\frac{\mathrm dx}{\ln^2\tan x+\pi^2}=\color{red}\frac{1}{\pi}-\color{blue}\frac{1}{4}$$

35.$$\int_{0}^{\pi/4}\frac{\mathrm dx}{\ln^2\tan^2(x)+\pi^2}=\color{red}\frac{\ln 2}{4\pi}$$

36.$$\int_{0}^{\pi/4}\frac{\mathrm dx}{(\ln^2\tan x+\pi^2)^2}=\color{red}\frac{1}{\pi^3}(3-C)-\color{purple}\frac{1}{8\pi^2}$$

37.$$\int_{0}^{\pi/4}\ln(\ln^2\tan x+\alpha^2\pi^2)=\pi\ln\left[\sqrt{2\pi}\frac{\Gamma\left(\frac{\alpha}{2}+\frac{3}{4}\right)}{\Gamma\left(\frac{\alpha}{2}+\frac{1}{4}\right)}\right]$$

38.$$\int_{0}^{1}\int_{0}^{1}\frac{x}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\alpha+xy}=\ln\left(\color{green}1+\frac{1}{\alpha}\right)$$

39.$$\int_{0}^{1}\int_{0}^{1}\frac{x}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{(\alpha+xy)^2}=\color{red}\frac{1}{\alpha+\alpha^2}$$

40.$$\int_{0}^{1}\int_{0}^{1}\frac{x}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{(\alpha+xy)^3}=\frac{2n+1}{8}\color{red}\prod_{j=1}^{n-1}\left(\frac{j}{j+2}\right)^2$$

41.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln^{n}(x)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\alpha-xy}=\color{green}\frac{\Gamma(n+1)}{1-\alpha}\left[\color{blue}\zeta(n+2)-\color{red}\operatorname{Li}_{n+2}\left(\frac{1}{\alpha}\right)\right]$$

42.$$\int_{0}^{1}\int_{0}^{1}\frac{xy}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{2-xy}=\color{green}\ln^2(2)$$

43.$$\int_{0}^{1}\int_{0}^{1}\frac{1-2xy}{2-xy}\cdot \frac{\ln x}{1-x^2y^2}\mathrm dx \mathrm dy=\color{green}\frac{\pi^2\ln 2-2\ln^3(2)-9\zeta(3)}{12}$$

44.$$\int_{0}^{\infty}\frac{x-\lfloor x \rfloor}{1+x^2}\mathrm dx=\ln 2$$

45.$$\int_{1}^{\infty}\frac{x-\lfloor x \rfloor}{x+x^2}\mathrm dx=\color{red}2(2\gamma-1)$$

46.$$\int_{0}^{\infty}\frac{\mathrm dx}{(x+\sqrt{\alpha+x^2})^n}=\color{red}\frac{n}{n^2-1}\cdot \color{blue}\frac{1}{\sqrt{\alpha^{n-1}}}$$

47.$$\int_{0}^{\infty}\frac{1}{\left(x+\sqrt{1+x^2}\right)^{2n+1}}\cdot \frac{\mathrm dx}{1+x^2}=(-1)^{n+1}\color{red}{(\color{blue}H_n^{*}-\ln 2)}$$,$$n\ge0$$

48.$$\int_{0}^{\infty}\frac{1}{\left(x+\sqrt{1+x^2}\right)^{2n}}\cdot \frac{\mathrm dx}{1+x^2}=G(n)$$,$$n\ge0$$

49.$$\int_{0}^{\infty}\frac{1}{\left(x+\sqrt{1+x^2}\right)^{2}}\cdot \frac{\mathrm dx}{1+x^2}=\color{green}2-\color{red}\frac{\pi}{2}$$

50.$$\int_{0}^{\infty}\frac{1}{\left(x+\sqrt{1+x^2}\right)^{4}}\cdot \frac{\mathrm dx}{1+x^2}=-\color{blue}\frac{2\cdot 2}{1\cdot 3}+\color{red}\frac{\pi}{2}$$

51.$$\int_{0}^{\infty}\frac{\ln x}{x+\sqrt{1+x^2}}\cdot \frac{\mathrm dx}{1+x^2}=\color{red}-\frac{\eta(2)}{2}-\color{purple}\frac{\ln^2(2)}{2}$$

52.$$\int_{0}^{\infty}\frac{\ln x}{x+\sqrt{1+x^2}}\cdot \frac{\mathrm dx}{(1+x^2)^2}=-\ln 2$$

53.$$\int_{0}^{\infty}\frac{\ln^2(x)}{x+\sqrt{1+x^2}}\cdot \frac{\mathrm dx}{(1+x^2)^2}=\color{orange}\lambda(2)+\ln^2(2)$$

54.$$\int_{0}^{\infty}\frac{\ln x}{x+\sqrt{1+x^2}}\cdot \frac{\mathrm dx}{\sqrt{1+x^2}}=\color{purple}\ln 2-1$$

55.$$\int_{0}^{\infty}\frac{\ln x}{(x+\sqrt{1+x^2})^2}\cdot \frac{\mathrm dx}{1+x^2}=\color{brown}2(\ln 2-1)$$

56.$$\int_{0}^{\infty}\frac{x\ln x}{x+\sqrt{1+x^2}}\cdot \frac{\mathrm dx}{1+x^2}=\color{green}1-\ln 2$$

57.$$\int_{0}^{\infty}\frac{e^{-x}}{x+\sqrt{1+x^2}}\cdot \frac{\mathrm dx}{1+x^2}=\color{red}\frac{\zeta(2)}{4}$$

58.$$\int_{0}^{\pi}\sin x\ln^2\left(\cot \frac{x}{2}\right)\mathrm dx=\color{orange}\zeta(2)$$

59.$$\int_{0}^{\pi}\sin(2x)\ln\left(\cot \frac{x}{2}\right)\mathrm dx=\color{purple}2$$

60.$$\int_{0}^{\pi}\cos(2x)\ln^2\left(\cot \frac{x}{2}\right)\mathrm dx=\color{blue}2\pi$$

61.$$\int_{0}^{\pi}\cos(2x)\ln^4\left(\cot \frac{x}{2}\right)\mathrm dx=\color{red}3\pi^3$$

62.$$\int_{0}^{\pi/2}\cos(2x)\ln^3\left(\cot \frac{x}{2}\right)\mathrm dx=\color{purple}12C$$

63.$$\int_{0}^{1/\alpha}\frac{\arcsin(x)}{\sqrt{1+x}}\mathrm dx=2\cdot \color{red}\frac{2\sqrt{\alpha^2-1}+(1+\alpha)\arcsin\left(\frac{1}{\alpha}\right)}{\color{green}\sqrt{\alpha(1+\alpha)}}-4$$

64.$$\int_{0}^{1/\alpha}\frac{\arcsin(x)}{(1+x)^2}\mathrm dx=1-\color{blue}\frac{\sqrt{\alpha^2-1}+\alpha\arcsin\left(\frac{1}{\alpha}\right)}{1+\alpha}$$

65.$$\int_{0}^{1/\alpha}\frac{x\arcsin(x)}{(1-x^2)^2}\mathrm dx=\color{blue}\frac{\sqrt{\alpha^2-1}-\alpha^2\arcsin\left(\frac{1}{\alpha}\right)}{2-2\alpha^2}$$

66.$$\int_{0}^{1/\alpha}\frac{x\arcsin^2(x)}{\sqrt{1-x^2}}\mathrm dx=\frac{2\arcsin\left(\frac{1}{\alpha}\right)-\sqrt{\alpha^2-1}\left[\arcsin^2\left(\frac{1}{\alpha}\right)-2\right]}{2}-2$$

67.$$\int_{0}^{1/\sqrt{2}}\frac{x\arcsin^2(x)}{(1-x^2)^2}\mathrm dx=\left(\frac{\pi}{4}\right)^2-\frac{\pi}{4}+\frac{\ln 4}{4}$$

68.$$\int_{0}^{\infty}\sin(x\sinh\alpha)\sin(x\cosh\alpha)\ln(x)\frac{\mathrm dx}{x}=\color{purple}-\alpha \gamma$$

69.$$\int_{0}^{\infty}\sin(x\sinh\alpha)\cos(x\cosh\alpha)\ln(x)\frac{\mathrm dx}{x}=\color{purple}-\alpha\cdot \frac{\pi}{2}$$

70.$$\int_{0}^{\pi/2}\cos(2x)\ln\cot\left(\frac{x}{2}\right)\mathrm dx=\color{red}1$$

71.$$\int_{0}^{\pi/2}\cos^2(2x)\ln\cot\left(\frac{x}{2}\right)\mathrm dx=C+\color{red}\frac{1}{6}$$

72.$$\int_{0}^{\pi/2}\cos^2(x)\ln\cot\left(\frac{x}{2}\right)\mathrm dx=C+\color{red}\frac{1}{2}$$

73.$$\int_{0}^{\pi/2}\cos^2\left(\frac{x}{2}\right)\ln\cot\left(\frac{x}{2}\right)\mathrm dx=C+\color{red}\frac{\pi}{4}$$

74.$$\int_{0}^{\pi/2}\cos(2x)\ln\left(1+\cot\left(\frac{x}{2}\right)\right)\mathrm dx=\color{red}\frac{\pi}{4}$$

75.$$\int_{0}^{1}\frac{\ln x}{\sqrt{x}}\cdot \ln\left(2\cdot\frac{1-\sqrt{x}}{2-\sqrt{x}}\right)\mathrm dx=\color{red}4\ln(2)-4\ln^2(2)$$

75.$$\int_{0}^{1}\frac{\ln x}{\sqrt{x}}\cdot \ln\left(1-\frac{1}{2}\sqrt{x}\right)\mathrm dx=8-4\zeta(2)+4\ln^2(2)-4\ln(2)$$

76.$$\int_{0}^{1}\frac{\ln x}{\sqrt[n]{x}}\cdot \ln\left(2-\sqrt[n]{x}\right)\mathrm dx=G(n)$$

77.$$\int_{0}^{\pi}x{\ln(1+\sin x)}\frac{\mathrm dx}{\sin x}=\color{red}\frac{\pi^3}{8}$$

78.$$\int_{0}^{\infty}\frac{(x^2-3x+1)}{e^x}\ln^3(x) \mathrm dx=\color{red}-6\gamma$$

1.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{1+\left(\frac{\tan x}{\tan y}\right)^2}=\color{red}\frac{\pi^2}{8}$$

2.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\left(\frac{\sin x}{\sin y}\right)^2+\left(\frac{\cos x}{\cos y}\right)^2}=\color{blue}\frac{\pi}{4}$$

3.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\left(\left(\frac{\sin x}{\sin y}\right)^2+\left(\frac{\cos x}{\cos y}\right)^2\right)^2}=\color{green}\frac{\pi}{8}$$

4.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\frac{\sin x}{\sin y}+\frac{\cos x}{\cos y}\cdot \frac{\tan y}{\tan x}}=\color{purple}\frac{\pi}{4}$$

5.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\left(\frac{\sin x}{\sin y}+\frac{\cos x}{\cos y}\cdot \frac{\tan y}{\tan x}\right)^2}=\color{purple}\frac{\pi}{8}$$

6.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\left(\frac{\tan x}{\sin y}\right)^2+\left(\frac{\sin x}{\cos y}\right)^2}=-\color{blue}\frac{\pi}{15}$$

7.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin^2(x)}{1+\frac{\cos x}{\cos y}}\mathrm dx \mathrm dy=\color{red}\frac{\pi^2}{8}-\frac{1}{2}$$

8.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin^2(x)}{1+\left(\frac{\cos x}{\cos y}\right)^2}\mathrm dx \mathrm dy=\color{red}\frac{\pi}{4}$$

9.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin(x)}{1+\left(\frac{\cos x}{\cos y}\right)^2}\mathrm dx \mathrm dy=\color{blue}\pi\left(1-\frac{1}{\color{brown}\sqrt{2}}\right)$$

10.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin(x)}{\frac{\sin x}{\sin y}+\left(\frac{\cos x}{\cos y}\right)^2}\mathrm dx \mathrm dy=\color{blue}\pi\left(\color{brown}\ln 2-\frac{1}{2}\right)$$

11.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\left(\frac{\tan x}{\sin y}\right)^2+\left(\frac{1}{\cos y}\right)^2}=\color{blue}\frac{\pi}{8}$$

12.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\frac{\tan x}{\tan y}+\frac{\tan y}{\tan x}}=\color{blue}\frac{\pi^2}{16}$$

13.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\cos y\tan x+\cos x \tan y}=\color{blue}2C$$

14.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\frac{\cos y}{\cos x}+\frac{\tan y}{\tan x}}=\color{blue}\frac{\pi^2}{12}$$

15.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{\frac{\sin x}{\sin y}+\frac{\tan y}{\tan x}}=\color{orange}\frac{\pi^2}{12}$$

16.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin x}{(\cos y\tan x+\cos x \tan y)^2}\mathrm dx \mathrm dy=\color{purple}\frac{\pi}{4}$$

17.$$\int_{0}^{1}\int_{0}^{1}\frac{x}{x+y}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{(1-x)(1-y)}}=\color{red}2$$

18.$$\int_{0}^{1}\int_{0}^{1}\frac{x}{(x+y)^2}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{(1-x)(1-y)}}=\color{green}2C$$

19x.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{x+y}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{(1-x^2)(1-y^2)}}=\color{brown}\frac{\pi^2}{4}$$

20.$$\int_{0}^{1}\int_{0}^{1}\frac{x}{x+y}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{(1-x^2)(1-y^2)}}=\color{blue}\frac{\pi^2}{8}$$

21.$$\int_{0}^{1}\int_{0}^{1}\frac{x}{(x+y)^2}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{(1-x^2)(1-y^2)}}=\color{purple}\frac{\pi^2}{8}$$

22.$$\int_{0}^{1}\int_{0}^{1}\frac{x}{1-x^2-y^2}\mathrm dx \mathrm dy=\color{red}\ln 2$$

23.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{\cos^2(x)}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{red}2C-1$$

24.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{\sin^2(x)}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{blue}1$$

25.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{\cos(2x)}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{green}2(C-1)$$

26.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{\sin(2x)}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{brown}\pi-2$$

27.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{1}{\sin^2(x)}\cdot\frac{\mathrm dx\mathrm dy}{\sqrt{1-y^2\sin^2(x)}}=\color{brown}C-\frac{1}{2}$$

28.$$\int_{0}^{1}\int_{0}^{\pi/2}\sin x\tan x\cdot\frac{\mathrm dx\mathrm dy}{\sqrt{1-y^2\sin^2(x)}}=-\color{brown}\pi \ln 2$$

29.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{y^2}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{brown}\frac{2C+1}{4}$$

30.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{y^4}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{grey}\frac{18C+13}{64}$$

31.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{y}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{green}1$$

32.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{y\sin x}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{green}\ln 2$$

33.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{y\sin^2(x)}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{brown}\frac{\pi-2}{2}$$

34.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{y\cos(x)}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{blue}\frac{\pi-2}{2}$$

35.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{y}{\sin^2(x)}\cdot\frac{1}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{blue}\frac{1}{3}$$

36.$$\int_{0}^{1}\int_{0}^{\pi/2}\frac{y^2}{\sin^2(x)}\cdot\frac{1}{\sqrt{1-y^2\sin^2(x)}}\mathrm dx\mathrm dy=\color{RED}\frac{6C-1}{16}$$

37.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{y\left(\frac{1}{x}-1\right)}}=2\pi$$

38.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{y\left(\frac{1}{x}-y\right)}}=\pi$$

39.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{x(1-x)}}=2\pi$$

40.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{\frac{1-x}{y}}}=4$$

41.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}\cdot\frac{\mathrm dx \mathrm dy}{\sqrt{\frac{1-xy}{y}}}=2(\pi-2)$$

42.$$\int_{0}^{1}\int_{0}^{1}\frac{\mathrm dx \mathrm dy}{\sqrt{xy(1-xy)(1-x)}}=\color{purple}8C$$

43.$$\int_{0}^{1}\int_{0}^{1}\frac{y}{\sqrt{xy(1-xy)(1-x)}}\mathrm dx \mathrm dy=\color{green}2C+1$$

44.$$\int_{0}^{1}\int_{0}^{1}\frac{y^2}{\sqrt{xy(1-xy)(1-x)}}\mathrm dx \mathrm dy=\color{blue}\frac{18C+13}{16}$$

45.$$\int_{0}^{1}\int_{0}^{1}\frac{x}{\sqrt{xy(1-xy)(1-x)}}\mathrm dx \mathrm dy=\color{red}4$$

46.$$\int_{0}^{1}\int_{0}^{1}\frac{(1-x^{\alpha})(1-y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{purple}\frac{\ln \alpha!}{\alpha(\alpha-1)}-\color{green}\gamma$$, $$\alpha\ge2$$

47.$$\int_{0}^{1}\int_{0}^{1}\frac{(1+x^{\alpha})(1-y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{red}\frac{\ln \alpha!}{\alpha(\alpha-1)}+\color{blue}\gamma$$, $$\alpha\ge2$$

48.$$\int_{0}^{1}\int_{0}^{1}\frac{(1-x)(1-y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=2\color{blue}\gamma-1$$

49.$$\int_{0}^{1}\int_{0}^{1}\frac{(1+x)(1-y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{red}-1$$

50.$$\int_{0}^{1}\int_{0}^{1}\frac{(1-x)(1+y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{red}-1$$

51.$$\int_{0}^{1}\int_{0}^{1}\frac{x(1-y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{red}\gamma-1$$

52.$$\int_{0}^{1}\int_{0}^{1}\frac{x^2(1-y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{purple}\frac{1}{2}\ln 2$$

53.$$\int_{0}^{1}\int_{0}^{1}\frac{x(1-y^2)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{orange}\ln 2$$

54.$$\int_{0}^{1}\int_{0}^{1}\frac{xy(1-y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=1-\color{red}\gamma -\color{orange}\ln 2$$

55.$$\int_{0}^{1}\int_{0}^{1}\frac{(1+x)(1-y)}{1+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=1-2\color{orange}\ln 2$$

56.$$\int_{0}^{1}\int_{0}^{1}\frac{(1+x^2)(1-y)}{1+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\frac{1}{2}\color{red}\ln 2$$

57.$$\int_{0}^{1}\int_{0}^{1}\frac{x(1-y)}{1+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=1-\color{green}\ln \pi$$

58.$$\int_{0}^{1}\frac{1}{(1+x)^2}\sum_{j=1}^{\infty}x^{2^{j+1}-1}\mathrm dx=2\gamma-1$$

59.$$\int_{0}^{1}\int_{0}^{1}-\frac{1-x^{\alpha}y^{\beta}}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{red}\frac{1}{\alpha-\beta}\ln\left(\frac{\alpha!}{\beta!}\right)+\color{blue}\gamma$$

60.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln^n(x)\ln^m(y)}{1+x^2y^2}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=(-1)^{1+n+m}\color{blue}\frac{\beta(1+n+m)}{1+n+m}\cdot \color{red}\Gamma(n+1)\Gamma(m+1)$$

61.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{1-x^2y^2}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{blue}\beta^2(1)$$

62.$$\int_{0}^{1}\int_{0}^{1}\frac{xy}{(1+x^2y^2)^n}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{blue}\frac{1-2^{n-1}}{4}\cdot \color{red}\prod_{j=1}^{n-2}\frac{j}{2j+2}$$, $$n\ge2$$

63.$$\int_{0}^{1}\int_{0}^{1}\frac{xy}{1+\alpha x^2y^2}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{blue}-\frac{1}{2\alpha}\ln (\alpha+1)$$

64.$$\int_{0}^{1}\int_{0}^{1}\frac{xy}{\sqrt{1+\alpha x^2y^2}}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{green}\frac{1-\sqrt{\alpha+1}}{\alpha}$$

65.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{1+\alpha xy}}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{blue}\frac{2(\color{red}1-\sqrt{\alpha})}{\alpha}$$

66.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{1+4xy}}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{purple}-\frac{1}{\phi}$$

67.$$\int_{0}^{1}\int_{0}^{1}\frac{xy}{\sqrt{1+4x^2y^2}}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{green}\frac{1}{2\phi}$$

68.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{(2-xy)^n}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{blue}\frac{1-2^{n-1}}{2}\cdot \color{red}\prod_{j=1}^{n-2}\frac{j}{2j+2}$$, $$n\ge2$$

69.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{\alpha-xy}}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{blue}2(\sqrt{\color{green}\alpha-1}-\sqrt{\alpha})$$

70.$$\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{\alpha+xy}}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{blue}2(\sqrt{\color{green}\alpha+1}-\sqrt{\alpha})$$

71.$$\int_{0}^{1}\int_{0}^{1}\frac{1-2xy}{\alpha+xy}\cdot \frac{\mathrm dx \mathrm dy}{\alpha+1-xy}=\frac{1}{2}\color{blue}\ln^2\left(\color{red}\frac{\alpha+1}{\alpha}\right)$$

72.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(1-x)}{\sqrt{1-xy}}\mathrm dx \mathrm dy=-\color{blue}\ln\left(\color{red}4+\frac{1}{2}\cdot \frac{1}{\phi^2+1}\right)$$

73.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln^n(1-x)}{1-xy}\mathrm dx \mathrm dy=(-1)^n\color{blue}\Gamma(n+2)\color{red}\zeta(n+2)$$

74.$$\int_{0}^{1}\int_{0}^{1}\frac{(\ln x\ln y)^n}{1-xy}\mathrm dx \mathrm dy=\color{blue}\Gamma^2(n+1)\color{red}\zeta(2n+2)$$

75.$$\int_{0}^{1}\int_{0}^{1}\frac{x(1+2x)}{1+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{red}\ln\pi$$

76.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln x}{1+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{blue}\frac{\eta(2)}{2}$$

77.$$\int_{0}^{1}\int_{0}^{1}\frac{x\ln x}{1+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{green}\ln \pi-1$$

78.$$\int_{0}^{1}\int_{0}^{1}\frac{x\ln x}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{purple}1-\gamma$$

79.$$\int_{0}^{1}\int_{0}^{1}\frac{x\ln x\ln y}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{brown}1-2\gamma$$

80.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln^n(x)}{1+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=\color{brown}\frac{(-1)^n}{n+1}\cdot\color{blue}\Gamma(n)\eta(n)$$

81.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln^n(x)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=\color{red}\frac{(-1)^n}{n+1}\cdot\color{blue}\Gamma(n)\zeta(n)$$

82.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\phi-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\color{red}\ln\phi$$

83.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\phi^2-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\frac{1}{2}\color{red}\ln\phi$$

84.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\phi^2+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\frac{1}{2}\color{red}\ln\left(\color{blue}1+\frac{1}{\phi^2}\right)$$

85.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\phi+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\frac{1}{2}\color{blue}\ln\phi$$

86.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\frac{1}{\phi}+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\color{purple}\ln\phi$$

87.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\frac{1}{\phi^2}+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\frac{1}{2}\color{purple}\ln(\color{green}1+\phi^2)$$

88.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\frac{2}{n+\frac{1}{\phi}}+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\frac{1}{2}\color{brown}\ln\left(\color{green}\frac{n+\phi^2}{2}\right)$$

89.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\frac{1}{n+\frac{1}{\phi}}+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\frac{1}{2}\color{brown}\ln\left(\color{red}n+\phi\right)$$

90.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\phi^{n}+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\frac{1}{2}\color{brown}\ln\left(\color{red}1+\phi^{-n}\right)$$

91.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\phi^{n}-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\frac{1}{2}\color{red}\ln\left(\color{blue}1-\phi^{-n}\right)$$, $$n\ge 1$$

92.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{\frac{1}{n+\frac{1}{\phi^m}}+xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\frac{1}{2}\color{red}\ln\left(\color{red}1+n+\phi^{-m}\right)$$

93.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln x}{1+x^2y^2}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\color{red}\frac{\pi}{8}$$

94.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln x}{1+\alpha xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\color{red}\frac{1}{2\alpha}\ln(\color{blue}1+\alpha)$$

95.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln x}{1+\sqrt{xy}}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=\color{red}\ln 2-1$$

96.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln x}{(1+\sqrt{xy})^2}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=\color{green}\frac{1}{2}-\ln 2$$

97.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln x}{(1+\sqrt{xy})^3}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=\color{orange}\frac{1}{8}$$

98.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln x}{(1+\alpha\sqrt{xy})^3}\cdot \frac{\mathrm dx \mathrm dy}{\ln^2(xy)}=-\color{purple}\frac{1}{2(1+\alpha)^2}$$

99.$$\int_{0}^{1}\int_{0}^{1}\frac{\ln x}{(1+\sqrt{xy})^3}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{purple}\ln 2-\frac{1}{2}$$

1.$$\int_{0}^{\infty}\frac{e^{-x}\arctan(e^{-x})}{\cosh x}\mathrm dx=C-\color{red}\frac{\pi}{4}\cdot \color{blue}\ln 2$$

2.$$\int_{0}^{\infty}\frac{e^{-x}\arctan(e^{-x})}{1+\cosh x}\mathrm dx=\color{red}\frac{\pi-2}{2}\cdot \color{blue}\frac{\ln 2}{2}$$

3.$$\int_{0}^{1}\frac{1}{x}\ln\left(\frac{1-x^{\alpha}}{1-x^{\alpha+1}}\right)\mathrm dx=-\color{purple}\frac{\zeta(2)}{\alpha(1+\alpha)}$$

4.$$\int_{0}^{1}\frac{1}{x}\ln\left(\frac{1+x^{\alpha}}{1+x^{\alpha+1}}\right)\mathrm dx=\color{blue}\frac{\eta(2)}{\alpha(1+\alpha)}$$

5.$$\int_{0}^{1}\frac{\sqrt{\alpha+x^2}}{\alpha+x^2\sin^2(\beta)}\mathrm dx=\frac{\operatorname{arcsinh}\left(\frac{1}{\sqrt{\alpha}}\right)-\cos(\beta)\operatorname{arctanh}\left(\frac{\cos(\beta)}{\sqrt{1+\alpha}}\right)}{\sin^2(\beta)}$$

6.$$\int_{0}^{1}\frac{\sqrt{\alpha+x^2}}{\alpha+x^2\cos^2(\beta)}\mathrm dx=\frac{\operatorname{arcsinh}\left(\frac{1}{\sqrt{\alpha}}\right)-\sin(\beta)\operatorname{arctanh}\left(\frac{\sin(\beta)}{\sqrt{1+\alpha}}\right)}{\cos^2(\beta)}$$

7.$$\int_{0}^{1}\frac{x^{\alpha}-x^{\alpha-1}}{\ln(1+x)}\mathrm dx=\color{green}\sum_{k=0}^{\alpha-1}(-1)^{k+\alpha}\left(\color{purple}{\alpha-1 \choose k}+{\alpha \choose k}\right)\operatorname{li(2^{k+1})}+\color{blue}\operatorname{li(2^{\alpha+1})}+\color{red}(-1)^{\alpha}\ln F(\alpha)$$

8.$$\int_{0}^{1}\frac{x^{\alpha}+x^{\alpha+1}}{\ln(1+x)}\mathrm dx=\color{blue}\sum_{k=0}^{\alpha}(-1)^{k+\alpha}{\alpha \choose k}\color{purple}\operatorname{li(2^{k+2})}+\color{red}(-1)^{\alpha}\ln F(\alpha)$$

9.$$\int_{0}^{1}\frac{(x^{\alpha}+x^{\alpha+1})^{\beta}}{\ln(1+x)}\mathrm dx=\color{blue}\sum_{k=0}^{\alpha\beta}(-1)^{k+\alpha}{\alpha \choose k}\color{purple}\operatorname{li(2^{k+1+\beta})}+\color{red}(-1)^{\alpha\beta}\ln F(\alpha)$$

10.$$\int_{0}^{1}\frac{\cos(\alpha\ln(1-x))}{\ln(1-x)}\cdot x\mathrm dx=\frac{1}{2}\ln\left(\frac{\color{red}\alpha^2+1^2}{\alpha^2+2^2}\right)$$

11.$$\int_{0}^{1}\frac{\sin(\alpha\ln(1-x))}{\ln(1-x)}\cdot x\mathrm dx=\arctan\left(\frac{\color{red}\alpha}{\alpha^2+2}\right)$$

12.$$\int_{0}^{1}\frac{\arctan\left(\frac{\alpha x}{1-x}\right)}{\sqrt{x(1-x)}}\cdot \frac{\mathrm dx}{x}=\pi\color{red}\sqrt{2\alpha}$$

13.$$\int_{0}^{1}\frac{\arctan(\alpha x)}{\sqrt{x(1-x)}}\cdot \frac{\mathrm dx}{x}=\color{green}\alpha \pi\cdot \color{red}\sqrt{\frac{2}{1+\sqrt{1+\alpha^2}}}$$

14.$$\int_{0}^{1}\frac{\arctan\left(\frac{x}{1-x}\right)}{\sqrt{x(1-x)}}\mathrm dx=\color{blue}\left(\frac{\pi}{2}\right)^2$$

15.$$\int_{0}^{1}\frac{\ln(1+x)}{(\alpha+x)^2}\mathrm dx=\frac{1}{\alpha^2-1}\left(\color{blue}\alpha\ln\left(\frac{\alpha}{1+\alpha}\right)+\color{red}\ln\left(\frac{4\alpha}{1+\alpha}\right)\right)$$

16.$$\int_{0}^{1}\frac{\ln(x^2-x+1)}{x^2-x}\mathrm dx=\color{green}\left(\frac{\pi}{3}\right)^2$$

17.$$\int_{0}^{1}\frac{\ln(2x^2-2x+1)}{x^2-x}\mathrm dx=\color{purple}\left(\frac{\pi}{2}\right)^2$$

18.$$\int_{0}^{1}\frac{\ln(3x^2-3x+1)}{x^2-x}\mathrm dx=\color{blue}\left(\frac{2\pi}{3}\right)^2$$

19.$$\int_{0}^{1}\frac{\ln(4x^2-4x+1)}{x^2-x}\mathrm dx=\color{red}\pi^2$$

20.$$\int_{0}^{\infty}\frac{\coth^2(x)+\alpha}{(\coth^2(x)-x)+\frac{\pi^2}{4}}\mathrm dx=\color{green}\frac{6\alpha}{5}+\color{red}2$$

21.$$\int_{0}^{\infty}\frac{\coth^2(x)+\alpha}{(\coth^2(x)+x)+\frac{\pi^2}{4}}\mathrm dx=\color{red}\frac{\alpha}{2}+\color{blue}1$$

22.$$\int_{0}^{\infty}\frac{\coth^2(x)+\alpha-1}{(\coth^2(x)-\alpha x)+\frac{\pi^2}{4}}\mathrm dx=\color{red}2$$

23.$$\int_{-\infty}^{+\infty}\frac{x\coth^5(x)}{(\coth x-x)^2+\frac{\pi^2}{4}}\cdot \frac{\mathrm dx}{(\coth x+x)^2+\frac{\pi^2}{4}}=\color{red}\frac{\pi^2}{4}-\color{blue}1$$

24.$$\int_{0}^{\infty}\frac{x^3+\frac{1}{x^3}}{x^5+\frac{1}{x^5}}\mathrm dx=\color{blue}\frac{\pi}{\sqrt{5}}$$

25.$$\int_{0}^{\infty}\frac{\cos\left(\frac{x}{\alpha}\right)-1}{x^2}\cdot \cos x \ln x \mathrm dx=\color{purple}\frac{\pi}{4\alpha}\color{red}\ln\left(\color{blue}\frac{(\alpha-1)^{\alpha-1}\cdot (\alpha+1)^{\alpha+1}}{\alpha^{2\alpha}}\right)$$

26.$$\int_{0}^{\infty}\frac{\cos\left(\frac{x}{\alpha}\right)-1}{x}\cdot \cos x \ln x \mathrm dx=\frac{\color{red}\ln^2\left(\frac{\alpha}{\alpha-1}\right)+\color{blue}\ln^2\left(\frac{\alpha}{\alpha+1}\right)+\color{green}2\gamma\ln\left(\frac{\alpha^2-1}{\alpha^2}\right)}{4}$$

27.$$\int_{0}^{\infty}\frac{\sin\left(\frac{x}{\alpha}\right)-1}{x}\cdot \sin x \ln x \mathrm dx=\color{red}G(\alpha)$$

28.$$\lim_{n\to \infty}\int_{1/n}^{n}\left(\frac{1}{2}-\cos\left(\frac{x}{\alpha}\right)\right)\frac{\mathrm dx}{x}=\color{green}\gamma-\ln\alpha$$

29.$$\lim_{n\to \infty}\int_{1/n}^{n}\left(x-\cos\left(\frac{x}{\alpha}\right)\right)\frac{\mathrm dx}{x}=\lim_{n\to \infty}\int_{1/n}^{n}\left(x-\cos^2\left(\frac{x}{\alpha}\right)\right)\frac{\mathrm dx}{x}=\color{red}\frac{\pi}{2\alpha}$$

30.$$\lim_{n\to \infty}\int_{1/n}^{n}\left(1-\cos x\right)\frac{\mathrm dx}{x^2}=\color{blue}\frac{\pi}{2}$$

31.$$\int_{0}^{\infty}\left(\frac{1}{\sqrt{1+(\alpha x)^2}}-\cos x\right)\frac{\mathrm dx}{x}=\gamma-\color{blue}\ln\left(\frac{\alpha}{2}\right)$$

32.$$\int_{0}^{\infty}\frac{\cos^2(x)-\cos^2(x^2)}{x^2}\mathrm dx=\color{red}\frac{\sqrt{\pi}-\pi}{2}$$

33.$$\int_{0}^{\pi/2}\sin x\arccot(\sqrt{\cos x})\mathrm dx=\color{red}1$$

34.$$\int_{0}^{\pi/2}\sin^3(x)\arccot(\sqrt{\cos x})\mathrm dx=\color{blue}1-\frac{1}{3}\cdot \color{red}\left(1-\frac{1}{3}+\frac{1}{5}\right)$$

35.$$\int_{0}^{\pi/2}\sin^5(x)\arccot(\sqrt{\cos x})\mathrm dx=\color{blue}1-\color{red}\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{5}\right)+\color{purple}\frac{1}{5}\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\right)$$

36.$$\int_{0}^{\pi/2}\sin^{2n-1}(x)\arccot(\sqrt{\cos x})\mathrm dx=\color{blue}G(n)$$, $$n\ge1$$

36.$$\int_{0}^{\pi/\alpha}\sin(x)\arccot(\sqrt{\cos x})\mathrm dx=1-\color{red}\sqrt{\cos\left(\frac{\pi}{\alpha}\right)}+\color{blue}\arctan\sqrt{\cos\left(\frac{\pi}{\alpha}\right)}-\color{pruple}\cos\left(\frac{\pi}{\alpha}\right)\arccot\sqrt{\cos\left(\frac{\pi}{\alpha}\right)}$$

37.$$\int_{0}^{1}\frac{\arccos(x^2)}{1+2x^2}\cdot \frac{\mathrm dx}{\sqrt{1+x^2}}=\color{red}\frac{\pi^2}{12}$$

38.$$\int_{0}^{1}\frac{\arccos(x^2)}{(1+2x^2)^2}\cdot \frac{\mathrm dx}{\sqrt{1+x^2}}=\color{red}\frac{\pi}{6}\left(\color{blue}3-\sqrt{3}\right)$$

39.$$\int_{0}^{1}\frac{1}{\sqrt{1+x^2}}\cdot \arccos(x^2)\mathrm dx=\color{orange}\frac{\pi}{2}\ln 2$$

40.$$\int_{0}^{1}\frac{(1+2x^2)^3}{\sqrt{1+x^2}}\cdot \arccos(x^2)\mathrm dx=\color{purple}\frac{17\pi}{12}$$

41.$$\int_{0}^{1}\frac{x^2}{\sqrt{1+x^2}}\cdot \arccos(x^2)\mathrm dx=\color{orange}\frac{\pi}{4}\left(1-\ln 2\right)$$

42.$$\int_{0}^{1}\frac{\arctan\sqrt{\alpha+x}}{\sqrt{\alpha+x}}\mathrm dx=\color{red}\ln\left(\frac{1+\alpha}{2+\alpha}\right)-\color{blue}2\sqrt{\alpha}\arctan\sqrt{\alpha}+\color{green}2\sqrt{1+\alpha}\arctan\sqrt{1+\alpha}$$

43.$$\int_{0}^{\pi/2}\ln\left[x^2+\ln^2\left(\frac{\alpha}{\beta}\cos x\right)\right]\mathrm dx=\color{blue}\pi\ln\ln\color{red}\left(\frac{2\beta}{\alpha}\right)$$, $$\alpha\le\beta$$

44.$$\int_{0}^{\pi/2}\ln\left[x^2+\ln^2\left(\frac{3}{2}\cos x\right)\right]\mathrm dx=\color{green}\pi\ln\ln\color{red}\left(\frac{4}{3}\right)^3$$

45.$$\int_{0}^{\pi/2}\ln\left[x^2+\ln^2\left(\frac{4}{3}\cos x\right)\right]\mathrm dx=\color{red}\pi\ln\ln\color{purple}\left(\frac{3}{2}\right)^2$$

46.$$\lim_{n\to \infty}\int_{1/n}^{n}\left(\cos x-\cos\left(\frac{x}{2}\right)\right)\frac{\ln x}{x}\mathrm dx=2\gamma-1$$

47.$$\lim_{n\to \infty}\int_{1/n}^{n}\left(\frac{1}{2}-\frac{\sin x}{x}\right)\frac{\mathrm dx}{x}=\color{red}\gamma-1$$

48.$$\lim_{n\to \infty}\int_{1/n}^{n}\left(\cos x-\frac{\sin x}{x}\right)\frac{\mathrm dx}{x}=-\color{red}1$$

49.$$\int_{0}^{\pi/4}\frac{\mathrm dx}{\cos x\sqrt{\cos(2x)}}=\color{red}\frac{\pi}{2}$$

50.$$\int_{0}^{\pi/4}\frac{\mathrm dx}{\cos^{2n-1}x\sqrt{\cos(2x)}}=\color{blue}G(n)$$, $$n\ge1$$

51.$$\int_{0}^{\pi/4}\frac{\sin^{n}(x)}{\cos x}\sqrt{\cos(2x)}\mathrm dx=-\frac{\pi}{4}+\color{red}1-\frac{1}{6}\sum_{j=1}^{n-1}\color{purple}\prod_{k=1}^{j-1}\frac{k}{2k+3}=-\frac{\pi}{4}+\color{blue}{1-\frac{1}{6}\left(1+\frac{1}{5}+\frac{1\cdot 2}{5\cdot 7}+\frac{1\cdot2\cdot 3}{5\cdot 7\cdot 9}+\cdots\right)}$$, $$n\ge1$$

52.$$\int_{0}^{\pi/4}\frac{\sin^{n}(x)}{\cos x}\frac{\mathrm dx}{\sqrt{\cos(2x)}}=\color{red}\frac{\pi}{4}-\frac{1}{2}\sum_{j=1}^{n-1}\color{purple}\prod_{k=1}^{j-1}\frac{k}{2k+1}=\color{blue}{\frac{\pi}{4}-\frac{1}{2}\left(1+\frac{1}{3}+\frac{1\cdot 2}{3\cdot 5}+\frac{1\cdot2\cdot 3}{3\cdot 5\cdot 7}+\cdots\right)}$$, $$n\ge1$$

53.$$\int_{0}^{\pi/4}\frac{\mathrm dx}{\cos x\sqrt{\alpha+\cos(2x)}}=\color{red}\frac{1}{\sqrt{\alpha-1}}\cdot \color{red}\operatorname{arctanh}\left(\color{purple}\sqrt{\frac{\alpha-1}{2\alpha}}\right)$$

54.$$\int_{0}^{\pi/4}\frac{\tan x}{\cos x\sqrt{\alpha+\cos(2x)}}\mathrm dx=\frac{\sqrt{2\alpha}-\sqrt{\alpha+1}}{\alpha-1}$$

55.$$\int_{0}^{\pi/4}\frac{\tan x}{\sqrt{\alpha+\cos(2x)}}\mathrm dx=\color{red}\frac{\operatorname{arctanh}\left(\sqrt{\frac{\alpha}{\alpha-1}}\right)-\color{blue}\operatorname{arctanh}\left(\sqrt{\frac{\alpha+1}{\alpha}}\right)}{\sqrt{\alpha-1}}$$

56.$$\int_{0}^{1}\int_{0}^{1}\frac{(x-1)^2}{(1-xy)\ln(xy)}\mathrm dy\mathrm dx=\color{red}\frac{1}{2}\ln 2-\color{purple}\gamma$$

57.$$\int_{0}^{\infty}\left(\frac{1}{(1+x)^n}-\frac{\sin x}{x}\right)\frac{\mathrm dx}{x}=\color{purple}\gamma-1-\color{green}H_{n}$$

58.$$\int_{0}^{\infty}\left(\frac{1}{1+x^n}-\frac{\sin x}{x}\right)\frac{\mathrm dx}{x}=\color{purple}\gamma-1$$

59.$$\int_{0}^{\infty}\left(\frac{1}{(1+x)^n}+\frac{e^{-x}-1}{x}\right)\frac{\mathrm dx}{x}=\color{red}\gamma-1-\color{green}H_{n}$$

60.$$\int_{0}^{\infty}\left(\frac{1}{1+x^n}+\frac{e^{-x}-1}{x}\right)\frac{\mathrm dx}{x}=\color{brown}\gamma-1$$

61.$$\int_{0}^{\infty}\left(\frac{\cos x-1}{x^2}+\frac{1}{2}\cdot \frac{1}{(1+x)^n}\right)\frac{\mathrm dx}{x}=\frac{\color{red}\gamma-H_2-\color{orange}H_n}{2}$$

62.$$\int_{0}^{\infty}\left[\frac{x}{\sinh\left(\frac{x}{2}\right)}+2e^{-\alpha x}(2e^{-\alpha x}-3)\right]\frac{\mathrm dx}{x}=2\color{red}\left(\color{blue}\gamma+\ln \alpha\right)$$

63.$$\int_{0}^{\infty}\left[\frac{x}{\sinh\left(\frac{x}{2}\right)}+e^{-\alpha x}(e^{-\alpha x}-3)\right]\frac{\mathrm dx}{x}=2\color{red}\left(\color{blue}\gamma+\ln \alpha\right)+\ln 8$$

64.$$\int_{0}^{\infty}\left[\frac{x}{\sinh\left(\frac{x}{2}\right)}+2e^{-\alpha x}(2e^{-\alpha x}-3)\right]\mathrm dx=\color{blue}\pi^2-\color{red}\frac{4}{\alpha}$$

65.$$\int_{0}^{\infty}\left[\frac{x}{\sinh\left(\frac{x}{2}\right)}+2e^{-\alpha x}(2e^{-\alpha x}-3)\right]\cdot x\mathrm dx=\color{blue}28\zeta(3)-\color{red}\frac{5}{\alpha^2}$$

66.$$\int_{0}^{\infty}\left[\frac{x}{\sinh\left(\frac{x}{2}\right)}+2e^{-\alpha x}(2e^{-\alpha x}-3)\right]\cdot x^2\mathrm dx=\color{blue}180\zeta(4)-\color{red}\frac{11}{2\alpha^3}$$

67.$$\int_{0}^{\infty}\left[\frac{x}{\sinh\left(\frac{x}{2}\right)}+e^{-x/2}(e^{-x/2}-3)\right]^2\frac{\mathrm dx}{x}=-4\color{red}\left(\color{blue}1+\gamma\right)+\color{red}\ln(\color{purple}2\cdot 3^6)$$

68.$$\int_{0}^{\infty}\left[\frac{x}{\sinh\left(\frac{x}{2}\right)}+e^{-\alpha x}(e^{-\beta x}-3)\right]\frac{\mathrm dx}{x}=2\color{red}\gamma+\color{blue}\ln\left(\color{green}\frac{16\alpha^3}{\alpha+\beta}\right)$$

69.$$\int_{0}^{\infty}\left[\frac{x}{\sinh\left(\frac{x}{2}\right)}+2e^{-\alpha x}(2e^{-\beta x}-3)\right]\frac{\mathrm dx}{x}=2\left[\color{red}\gamma+\color{blue}\ln\left(\color{green}\frac{4\alpha^3}{(\alpha+\beta)^2}\right)\right]$$

70.$$\int_{0}^{\infty}\left(\frac{1}{\sqrt{1+4x^{\alpha}}}-e^{-x}\right)\frac{\mathrm dx}{x}=\color{purple}\gamma$$

71.$$\int_{0}^{\infty}\left(\frac{1}{\sqrt{1+4x^2}}-e^{-x}\right)\frac{\ln x}{x}\mathrm dx=-\color{purple}\frac{\gamma^2}{2}$$

72.$$\int_{0}^{\infty}\left(\frac{1}{\sqrt{1+x^2}}-e^{-x}\right)\frac{\ln x}{x}\mathrm dx=\frac{1}{2}\color{red}(\ln 2-\gamma)\color{blue}(\ln 2+\gamma)$$

73.$$\int_{0}^{\infty}\left(\frac{e^{-nx}}{e^{x}-1}-\frac{1}{x}\right)e^{-mx}\mathrm dx=\gamma-H_{m+n}+\ln m$$

74.$$\int_{0}^{\infty}\left(\frac{e^{nx}}{e^{x}-1}-\frac{1}{x}\right)e^{-mx}\mathrm dx=\gamma-H_{m-n}+\ln m$$

75.$$\int_{0}^{\infty}\left(\frac{e^{x}}{e^{x}-1}-\frac{1}{x}\right)e^{-x}\mathrm dx=\gamma$$

76.$$\int_{0}^{\infty}\left(\frac{\cosh(nx)}{e^{x}-1}-\frac{1}{x}\right)e^{-mx}\mathrm dx=\color{purple}\gamma+\ln m-\color{red}\frac{H_{m+n}+H_{m-n}}{2}$$

77.$$\int_{0}^{\infty}\frac{\sinh(nx)}{e^{x}-1}\cdot e^{-mx}\mathrm dx=\color{red}\frac{H_{m+n}-H_{m-n}}{2}$$

78.$$\int_{0}^{1}\left(\frac{1}{\ln^2(x)}+\frac{(\alpha-1)x^{\alpha+1}-\alpha x^{\alpha}}{(1-x)^2}\right)\frac{\mathrm dx}{x}=\color{red}\frac{2\alpha-1}{\alpha}$$, $$\alpha\ge1$$

79.$$\int_{0}^{1}\left(\frac{1}{\ln^2(x)}+\frac{(\alpha-1)x^{\alpha+1}-\alpha x^{\alpha}}{(1-x)^2}\right)\cdot\ln(x)\mathrm dx=\color{orange}H_{\alpha}^2-H_{\alpha}+\color{purple}\gamma+1-\color{red}\zeta(2)$$, $$\alpha\ge1$$

80.$$\int_{0}^{1/n}\left(\frac{1}{\ln^2(x)}+\frac{(\alpha-1)x^{\alpha+1}-\alpha x^{\alpha}}{(1-x)^2}\right)\frac{\mathrm dx}{x}=\color{red}\frac{1}{\ln n}-\color{purple}\frac{1}{(n-1)n^{\alpha-1}}$$, $$\alpha\ge1$$

81.$$\int_{0}^{n}\left(\frac{1}{\ln^2(x)}+\frac{(\alpha-1)x^{\alpha+1}-\alpha x^{\alpha}}{(1-x)^2}\right)\frac{\mathrm dx}{x}=\color{purple}\frac{n^{\alpha}}{n-1}-\color{red}\frac{1}{\ln n}$$, $$\alpha\ge1$$

82.$$\int_{0}^{\frac{1}{2n}}\left(\frac{1}{\ln^2(x)}+\frac{1}{\alpha}\cdot \frac{\sqrt{x}(1+x)}{(1-x)^2}\right)\frac{\mathrm dx}{x}=\color{red}\frac{2\sqrt{2n}}{(2n-1)\alpha}+\frac{1}{\ln(2n)}$$

83.$$\int_{0}^{\frac{1}{2n+1}}\left(\frac{1}{\ln^2(x)}+\frac{1}{\alpha}\cdot \frac{\sqrt{x}(1+x)}{(1-x)^2}\right)\frac{\mathrm dx}{x}=\color{blue}\frac{\sqrt{2n+1}}{n\alpha}+\frac{1}{\ln(2n+1)}$$

84.$$\int_{0}^{1}\frac{2+\alpha x^2}{1+\alpha x^2}\cdot \frac{x}{\sqrt{1+\alpha x^2}}\cdot \arctan\left(\frac{\sqrt{1+\alpha x^2}}{x}\right)\mathrm dx=\color{blue}\frac{\arctan\sqrt{\alpha}}{\sqrt{\alpha}}$$

85.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\mathrm dx \mathrm dy}{(1+\cos x\cos y)^2}=\color{red}C+\color{blue}\frac{1}{2}$$

86.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\cos x}{1+\cos x\cos y}\mathrm dx \mathrm dy=\color{blue}\frac{\pi}{2}\ln 2$$

87.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\cos^2(x)}{1+\cos x\cos y}\mathrm dx \mathrm dy=\color{blue}2C-\color{brown}1$$

88.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\cos x\cos y}{1+\cos x\cos y}\mathrm dx \mathrm dy=\color{purple}\left(\frac{\pi}{2}\right)^2-2C$$

89.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{1}{1+\cos^2(x)\cos^2(y)}\mathrm dx \mathrm dy=\color{grey}\frac{\pi K\left(\frac{1}{2}\right)}{2\sqrt{2}}$$

90.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\cos x}{1+\cos^2(x)\cos^2(y)}\mathrm dx \mathrm dy=\color{green}\frac{\pi^2}{8}$$

91.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin^2(x)}{1+\cos x\cos y}\mathrm dx \mathrm dy=\color{brown}1$$

92.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin(x)}{1+\cos x\cos y}\mathrm dx \mathrm dy=\color{green}\frac{\pi^2}{8}$$

93.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin^2(x)}{1+\cos x\cos^2(y)}\mathrm dx \mathrm dy=\color{red}\frac{\pi}{3}$$

94.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin x \sin y}{1+\cos x\cos y}\mathrm dx \mathrm dy=\color{red}\frac{\pi^2}{12}$$

95.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin^2(x) \sin y}{1+\cos x\cos y}\mathrm dx \mathrm dy=1+\color{red}\frac{\pi}{2}\cdot \frac{\pi-4}{4}$$

96.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin^2(x) \sin^2(y)}{1+\cos x\cos y}\mathrm dx \mathrm dy=3-\color{red}\frac{\pi^2}{4}$$

97.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin x \sin y}{(1+\cos x\cos y)^2}\mathrm dx \mathrm dy=\ln 2$$

98.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\left(\frac{\sin x \sin y}{1+\cos x\cos y}\right)^2\mathrm dx \mathrm dy=\frac{\pi^2}{4}-2$$

99.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin x}{1+\frac{\cos x}{\cos y}}\mathrm dx \mathrm dy=\color{red}\frac{\pi}{2}-\ln 2$$

100.$$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin x\sin y}{1+\frac{\cos x}{\cos y}}\mathrm dx \mathrm dy=\color{green}\frac{1}{2}$$

-

1.$$\int_{-\infty}^{\infty}\frac{\arctan(x)-2\arctan(2x)+\arctan(3x)}{x^3}\mathrm dx=-\color{blue}{\pi}$$

2.$$\int_{-\infty}^{+\infty}\pi\cdot\frac{\cosh^2(2x)}{\cosh^2(\pi x)}\mathrm dx=\color{purple}1+\color{grey}\tan(1)+\color{green}\cot(1)$$

3.$$\int_{0}^{\pi/2}\frac{\tan(x)\cos(2x)}{\alpha+\tan(x)}\mathrm dx=\frac{\alpha}{(\alpha^2+1)^2}\left[\alpha^2+1-\alpha\pi+(1-\alpha^2)\ln(\alpha)\right]$$

4.$$\int_{0}^{\pi/2}\frac{\cos(2x)}{\alpha+\tan(x)}\mathrm dx=\frac{1}{(\alpha^2+1)^2}\left[\alpha^2+1-\alpha\pi+(1-\alpha^2)\ln(\alpha)\right]$$

5.$$\int_{0}^{\pi/2}\frac{\tan^2(x)\cos(2x)}{\alpha^2+\tan^2(x)}\mathrm dx=-\frac{\pi}{2}\cdot\frac{b}{(b+1)^2}$$

6.$$\int_{0}^{1}\ln\sin(\pi x)\ln^2[2\sin(\pi x)]\mathrm dx=-\color{red}\eta(1)\color{blue}\eta(2)-\color{green}\frac{1}{2}\eta(3)$$

7.$$\int_{-\infty}^{+\infty}\frac{\cos^{2k-1}(\alpha\arctan x)}{(1+x^2)^{\alpha/2}}\cdot\frac{\mathrm dx}{1+x^2}=\color{red}\frac{\pi}{2^{\alpha-1}}\cdot\color{blue}\frac{(2k-1)!!}{(2k)!!}$$

8.$$\int_{0}^{\infty}\frac{\cos(4\arctan x)}{1+x^2}\ln^2(x)\mathrm dx=\color{blue}\pi$$

9.$$\int_{0}^{\infty}\frac{\cos(3\arctan x)}{(1+x^2)^{3/2}}\ln^2(x)\mathrm dx=\color{blue}\frac{\pi}{2}$$

10.$$\int_{0}^{\infty}\frac{\cos(4\arctan x)}{(1+x^2)^{2}}\ln^2(x)\mathrm dx=\color{green}\frac{\pi}{2}$$

11.$$\int_{0}^{\infty}\frac{\cos(\arctan x)}{\sqrt{1+x^2}}\ln^2(x)\mathrm dx=\color{purple}\left(\frac{\pi}{2}\right)^3$$

12.$$\int_{0}^{\infty}\frac{\cos(4\arctan x)}{(1+x^2)^2}\ln^4(x)\mathrm dx=\color{grey}\frac{3\pi^3}{4}$$

13.$$\int_{0}^{1}\cosh^2(\alpha\arccos x)\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi}{4}+\frac{1}{4\alpha}\sinh(\alpha \pi)$$

14.$$\int_{0}^{1}\cosh(\alpha\arccos x)\cos(\alpha\operatorname{arccosh}x)\frac{\mathrm dx}{\sqrt{1-x^2}}=\color{green}\frac{\pi}{4}+\frac{1}{4\alpha}\sinh(\alpha \pi)$$

15.$$\int_{0}^{1}\sin(\alpha\operatorname{arccosh} x)\frac{\mathrm dx}{\sqrt{x^2-1}}=\color{green}\frac{2}{\alpha}\sinh^2\left(\frac{\alpha \pi}{4}\right)$$

16.$$\int_{0}^{1}\sin^2(\alpha\operatorname{arccosh} x)\frac{\mathrm dx}{\sqrt{1-x^2}}=-\color{blue}\frac{\pi}{4}+\color{gre}\frac{1}{4\alpha}\sinh(\alpha \pi)$$

17.$$\int_{0}^{1}\sinh(\alpha\arccos x)\sin(\alpha\operatorname{arccosh}x)\frac{\mathrm dx}{\sqrt{x^2-1}}=-\color{green}\frac{\pi}{4}+\frac{1}{4\alpha}\sinh(\alpha \pi)$$

18.$$\int_{0}^{1}\cos(\alpha x)\sinh(\operatorname{arccosh}x)\frac{\mathrm dx}{\sqrt{x^2-1}}=\color{green}\frac{1}{\alpha}\cdot \sin(\alpha)$$

19.$$\int_{0}^{1}\sin(\alpha x)\sinh(\operatorname{arccosh}x)\frac{\mathrm dx}{\sqrt{x^2-1}}=\color{green}\frac{2}{\alpha}\cdot \sin^2\left(\frac{\alpha}{2}\right)$$

20.$$\int_{0}^{1}\tan\left(\frac{\alpha} {x}\right)\sinh(\operatorname{arccosh}x)\frac{\mathrm dx}{\sqrt{x^2-1}}=-\color{purple}\alpha\ln\cos\left(\frac{1}{\alpha}\right)$$

21.$$\int_{0}^{1}\tan^2\left(\frac{\alpha} {x}\right)\sinh(\operatorname{arccosh}x)\frac{\mathrm dx}{\sqrt{x^2-1}}=\color{blue}\alpha\cot(\alpha)-1$$

22.$$\int_{0}^{1}\ln^{n}(x)\sinh(\operatorname{arccosh}x)\frac{\mathrm dx}{\sqrt{x^2-1}}=(-1)^n\color{brown}\Gamma(n+1)$$

23.$$\int_{0}^{1}\left(\frac{\arcsin x}{x\sqrt{1-x^2}}+2\frac{\ln(1+x^2)}{1+x^2}\right)\mathrm dx=\color{green}\pi\color{blue}\ln 2$$

24.$$\int_{0}^{\pi}\cos^{2n+1}(2x)\ln^2\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\mathrm dx=\color{blue}\pi^2\cdot\color{grey}\frac{(2n)!!}{(2n+1)!!}$$

24.$$n\ge 0$$

25.$$\int_{0}^{\pi}\cos^{2}(2x)\ln^2\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\mathrm dx=-\color{blue}\pi-\left(\frac{\pi}{2}\right)^3$$

26.$$\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}-\frac{1}{2}\right)\frac{\ln x}{(1-x)^2}\mathrm dx=-\color{brown}\frac{1}{4}$$

27.$$\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}-\frac{1}{2}\right)\frac{\ln^2 x}{1-x}\mathrm dx=\color{brown}\zeta(2)-\color{grey}\zeta(3)$$

28.$$\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}-\frac{1}{2}\right)\frac{\ln^2 x}{(1-x)^2}\mathrm dx=\color{brown}\frac{1}{2}$$

29.$$\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}-\frac{1}{2}\right)\frac{\ln^{2n+2} x}{(1-x)^2}\mathrm dx=\color{green}n(2n+1)!\zeta(2n+1)$$

29.$$n\ge 1$$

30.$$\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}-\frac{1}{2}\right)\frac{\ln^{2n} x}{1-x}\mathrm dx=\color{blue}\Gamma(2n)\left[\Gamma(2n)\zeta(2n)-n\zeta(2n+1)\right]$$

30.$$n\ge 1$$

31.$$\int_{0}^{\pi/2}x^2\sin^2(x)\sqrt{\frac{1-\cos x}{1+\cos x}}\mathrm dx=-\color{green}\frac{\pi-14}{4}\cdot\color{blue}\frac{\pi-2}{4}$$

32.$$\int_{0}^{\pi/2}\sin(x)\arctan^2\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\mathrm dx=\color{green}\frac{\pi+1}{2}\cdot\color{blue}\frac{\pi-2}{2}$$

33.$$\int_{0}^{\pi/2}\sin(x)\arctan^2(\cos x)\mathrm dx=\color{green}\beta(1)[\color{blue}\beta(1)+\ln 2]-\color{red}C$$

34.$$\int_{0}^{\pi/2}x^2\sin(x)\arctan^2\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\mathrm dx=\color{grey}3!-\frac{\pi^2}{2!}$$

35.$$\int_{-1}^{+1}\frac{\arctan(x)}{1-x}\ln\left(\frac{1+x^2}{2}\right)\mathrm dx=-\frac{\pi^3}{96}$$

36.$$\int_{-1}^{+1}\frac{\arctan(x)}{1-x}\ln\left(\frac{1+x}{2}\right)\mathrm dx=\frac{\pi^3}{104}$$

37.$$\int_{-1}^{+1}\frac{\arctan(x)}{1-x}\ln\left(\frac{1+x}{1+x^2}\right)\mathrm dx=\color{red}\left(\frac{\pi}{2}\right)^3\cdot\color{blue}\frac{12+13}{12\cdot13}$$

38.$$\int_{0}^{\pi/2}\frac{x^3}{3}\cdot \sin(4x)\arctan\left(\frac{1+\tan x}{1-\tan x}\right)\mathrm dx=\frac{\pi-0}{4}\cdot\frac{\pi+0}{4}\cdot \frac{\pi-1}{4}\cdot \frac{\pi+1}{4}$$

39.$$\int_{0}^{\pi/2}x^2\sin(4x)\frac{(1+\tan x)^2}{1-\tan x}\mathrm dx=\frac{\pi}{16}(\pi+1)(2\pi+7)-2$$

40.$$\int_{0}^{\pi/2}\frac{x}{\sin x}\ln\left(\frac{1+\cos x}{1-\cos x}\right)\mathrm dx=\color{blue}\left(\frac{\pi}{2}\right)^3$$

41.$$\int_{0}^{1}\frac{1-x^6}{1-x+x^2}\frac{\mathrm dx}{\ln x}=\color{green}-\ln(10)$$

42.$$\int_{0}^{1}\frac{1-x^{12}}{1+x+x^2}\frac{\mathrm dx}{\ln x}=-\color{red}\ln\left(\frac{22}{7}\right)$$

43.$$\int_{0}^{\pi/4}\frac{\mathrm dx}{\sin(2x)[\tan^{\alpha}(x)+\cot^{\alpha}(x)]^{2k-1}}=\frac{\pi}{2^{2k+1}\alpha}\cdot\frac{(2k-3)!!}{(2k-2)!!}$$

44.$$\int_{0}^{\pi/2}\frac{\mathrm dx}{\sin(2x)[\tan^{\alpha}(x)+\cot^{\alpha}(x)]^{2k-1}}=\frac{\pi}{2^{2k}\alpha}\cdot\frac{(2k-3)!!}{(2k-2)!!}$$

45.$$\int_{0}^{\pi/2}\frac{\mathrm dx}{\sin(2x)[\tan^{\alpha}(x)+\cot^{\alpha}(x)]^{2k}}=\frac{1}{4\alpha}\cdot\frac{\color{red}\Gamma^2(n)}{\color{green}\Gamma(2n)}$$

46.$$\int_{0}^{1}\arcsin\sqrt{1-x}\ln(x)\mathrm dx=-\color{blue}\frac{\pi}{2}\cdot\color{red}\ln 2$$

47.$$\int_{0}^{1}\arccos\sqrt{1-x}\ln(1-x)\mathrm dx=-\color{blue}\frac{\pi}{2}\cdot\color{red}\ln 2$$

48.$$\int_{0}^{1}\arccos\sqrt{1-x}\ln(x)\mathrm dx=\color{green}\frac{1}{2}\color{red}(\pi\ln 2-1)$$

49.$$\int_{0}^{1}\arcsin\sqrt{1-x}\ln(1-x)\mathrm dx=\color{green}\frac{\pi}{2}\color{red}(\pi\ln 2-1)$$

50.$$\int_{0}^{1}\arcsin\sqrt{1-x^2}\ln(x)\mathrm dx=\color{purple}\ln 2-2$$

51.$$\int_{0}^{1}\arccos\sqrt{1-x^2}\ln(x)\mathrm dx=\color{purple}2-\ln 2-\color{grey}\frac{\pi}{2}$$

52.$$\int_{0}^{1}\arcsin\sqrt{1-x}\ln^2(x)\mathrm dx=+\color{purple}\frac{\pi^3}{12}-\frac{\pi}{2}+\pi\ln^2(2)$$

53.$$\int_{0}^{1}\arccos\sqrt{1-x}\ln^2(1-x)\mathrm dx=+\color{purple}\frac{\pi^3}{12}-\frac{\pi}{2}+\pi\ln^2(2)$$

54.$$\int_{0}^{1}\arccos\sqrt{1-x}\ln^2(x)\mathrm dx=-\color{brown}\frac{\pi^3}{12}+3\frac{\pi}{2}-\pi\ln^2(2)$$

55.$$\int_{0}^{1}\arcsin\sqrt{1-x}\ln^2(1-x)\mathrm dx=-\color{brown}\frac{\pi^3}{12}+3\frac{\pi}{2}-\pi\ln^2(2)$$

56.$$\int_{0}^{1}\arcsin\sqrt{1-x}\ln(x-x^2)\ln\left(\frac{1}{x}-1\right)\mathrm dx=\pi\zeta(2)-2\pi[1-\ln^2(2)]$$

57.$$\int_{0}^{1}\arccos\sqrt{1-x}\ln(x-x^2)\ln\left(\frac{1}{x}-1\right)\mathrm dx=-\pi\zeta(2)+2\pi[1-\ln^2(2)]$$

58.$$\int_{0}^{1}\ln\left(1-\frac{x}{\alpha}\right)\frac{\mathrm dx}{x\sqrt{1-x^2}}=\frac{1}{2}\left(\pi-\arcsin\left(\frac{1}{\alpha}\right)\right)\arcsin\left(\frac{1}{\alpha}\right)$$

59.$$\int_{0}^{1}\frac{1-x(1-\ln x)}{x\ln^2(x)}\ln(1-x)\mathrm dx=-\color{orange}\gamma$$

60.$$\int_{0}^{\pi/4}\frac{1-\sin^2(x)}{1+\sin^2(x)}\ln\tan\left(\frac{x}{2}\right)\mathrm dx=-\frac{11}{9}$$

61.$$\int_{0}^{1}\frac{\ln^2(x)}{1+x+x^2}\mathrm dx=\color{green}\left(\frac{2}{3}\color{red}\pi\right)^3\color{blue}\frac{1}{3\sqrt{3}}$$

62.$$\int_{0}^{1}\frac{\ln^4(x)}{1+x+x^2}\mathrm dx=\color{green}\left(\frac{2}{3}\color{red}\pi\right)^5\color{purple}\frac{1}{\sqrt{3}}$$

63.$$\int_{0}^{1}\frac{\ln^6(x)}{1+x+x^2}\mathrm dx=\color{green}\left(\frac{2}{3}\color{red}\pi\right)^7\color{brown}\frac{7}{\sqrt{3}}$$

64.$$\int_{0}^{1}\frac{\ln(x)}{1+x+x^2}\mathrm dx=-\color{green}\sqrt{3}\cdot\color{grey}\frac{2}{3}Cl_2\left(\frac{2}{3}\color{blue}\pi\right)$$

65.$$\int_{0}^{\pi/2}\left[\tan(x)-\tan\left(\frac{x}{2}\right)\right]\ln\sin(x)\mathrm dx=\frac{1}{2}\color{blue}\ln^2(2)$$

66.$$\int_{0}^{\pi/2}\left[\tan(x)-2\tan\left(\frac{x}{2}\right)+\tan\left(\frac{x}{3}\right)\right]\ln\sin(x)\mathrm dx=\frac{5}{273}\color{blue}\pi^2$$

67.$$\int_{0}^{\pi/2}\frac{\sin(3x)\sin(4x)}{\sqrt{\sin(x)\sin(2x)}}\mathrm dx=\frac{188}{231}\color{blue}\sqrt{2}$$

68.$$\int_{0}^{1}\frac{\sin(3x)\sin(4x)}{\sqrt{\sin(x)\sin(2x)}}\mathrm dx=\frac{3\times 11}{4\times 10}$$

69.$$\int_{0}^{\pi/2}\sin^{2n}(x)\ln\sin(x)\mathrm dx=\color{red}\frac{\pi}{2}\left(\color{blue}H_{2n}^{*}-\ln 2\right)\cdot \color{purple}\frac{(2n-1)!!}{(2n)!!}$$

70.$$\int_{0}^{\pi/2}\cos^{2n}(x)\ln\sin(x)\mathrm dx=-\color{red}\frac{\pi}{2}\left(\color{green}\frac{H_{n}}{2}+\ln 2\right)\cdot \color{orange}\frac{(2n-1)!!}{(2n)!!}$$

71.$$\int_{0}^{\pi/2}\sin^{2n}(x) \tan^2 (x)\ln\sin(x)\mathrm dx=\color{red}\frac{\pi}{2}\left(\color{blue}\ln 2-H_{2n+1}^{*}\right)\cdot \color{pink}\frac{(2n+1)!!}{(2n-2)!!}$$

72.$$\int_{0}^{1}\arctan(x)\ln^{k}(x)\frac{\mathrm dx}{x}=(-1)^k\color{red}\Gamma(k+1)\color{blue}\beta(k+2)$$

73.$$\int_{0}^{\pi/2}\frac{\sin^2(x)(1-4\sin^2(x))}{\sqrt{1+\sin(2x)}}\mathrm dx=-\color{red}1$$

74.$$\int_{0}^{\pi}\frac{\sin^2(x)(1-4\sin^2(x))}{\sqrt{1+\sin(2x)}}\mathrm dx=-\color{green}\sqrt{2^3}$$

75.$$\int_{0}^{2\pi}\frac{\sin^2(x)(1-4\sin^2(x))}{\sqrt{1+\sin(2x)}}\mathrm dx=-\color{blue}\sqrt{2^5}$$

76.$$\int_{0}^{4\pi}\frac{\sin^2(x)(1-4\sin^2(x))}{\sqrt{1+\sin(2x)}}\mathrm dx=-\color{grey}\sqrt{2^7}$$

77.$$\int_{0}^{\pi/2}\frac{\sin^{2n-1}(2x)\cos(x)}{\sqrt{1+\sin(2x)}}\mathrm dx=\color{grey}\frac{1}{2}\frac{(2n-2)!!}{(2n-1)!!}$$

78.$$\int_{0}^{\pi/2}\frac{\sin^{2n}(2x)\cos(x)}{\sqrt{1+\sin(2x)}}\mathrm dx=\color{blue}\frac{\pi}{4}\frac{(2n-1)!!}{(2n)!!}$$

79.$$\int_{0}^{\pi/2}\frac{\sin^{2n-1}(2x)\cos(x)[1-4\sin^2(x)]}{\sqrt{1+\sin(2x)}}\mathrm dx=\frac{1}{2}\frac{(2n-2)!!}{(2n-1)!!}-\color{blue}\frac{\pi}{2}\frac{(2n-1)!!}{(2n)!!}$$

80.$$\int_{0}^{\pi/2}\frac{\sin^{2n}(2x)\cos(x)[1-4\sin^2(x)]}{\sqrt{1+\sin(2x)}}\mathrm dx=\frac{\pi}{4}\frac{(2n-1)!!}{(2n)!!}-\color{green}\frac{(2n)!!}{(2n+1)!!}$$

81.$$\int_{0}^{1}\int_{0}^{1}\ln(x)\ln(y)\frac{\mathrm dy \mathrm dy}{1-xy}=\color{blue}\zeta(4)$$

82.$$\int_{0}^{1}\int_{0}^{1}x\ln(x)\ln(y)\frac{\mathrm dy \mathrm dy}{1-xy}=2\color{blue}\zeta(2)-\color{green}3$$

83.$$\int_{0}^{1}\int_{0}^{1}\ln(x)\ln(y)\frac{\mathrm dy \mathrm dy}{(1-xy)^2}=\color{purple}\zeta(3)$$

84.$$\int_{0}^{1}\int_{0}^{1}\ln(x)\ln(y)\frac{\mathrm dy \mathrm dy}{(1+xy)^2}=\color{brown}\eta(3)$$

85.$$\int_{0}^{1}\int_{0}^{1}2\ln(x)\ln(y)\frac{\mathrm dy \mathrm dy}{(1-xy)^3}=\color{purple}\zeta(3)+\color{red}\zeta(2)$$

86.$$\int_{0}^{1}\int_{0}^{1}2\ln(x)\ln(y)\frac{\mathrm dy \mathrm dy}{(1+xy)^3}=\color{grey}\eta(3)+\color{red}\zeta(2)$$

87.$$\lim_{n \to \infty}\int_{0}^{\pi/2}\arctan[\tan^{2n}(\pi^2 x)]\mathrm dx=\color{blue}\frac{5}{4}$$

88.$$\int_{0}^{\infty}\frac{x^2-1}{x^2+1}\cdot\frac{\ln(1+x^3)}{x^2+1}\mathrm dx=\frac{3}{4}\color{red}\pi$$

89.$$\int_{0}^{\infty}\frac{2x-1}{(x+1)^4}\ln(1+x^3)\mathrm dx=\frac{2}{9}\cdot\color{red}\frac{\pi}{\sqrt{3}}$$

90.$$\int_{0}^{\infty}\ln(1+x^3)\frac{\mathrm dx}{(1+x)^2}=1+\frac{2}{9}\color{red}\pi\sqrt{3}$$

91.$$\int_{0}^{\pi/2}\frac{\ln\cos x}{\sin^2(x)}=-\frac{\pi}{2}$$

92.$$\int_{0}^{\pi/2}\frac{\ln^3(\cos x)}{\sin(x)}=-\color{red}\left(\frac{\pi}{2}\right)^4$$

93.$$\int_{0}^{\pi/2}\frac{\ln^2(\cos x)}{\sin(x)}=\color{red}\frac{7}{4}\zeta(3)$$

94.$$\int_{0}^{\pi/2}\frac{\ln^4(\cos x)}{\sin(x)}=\color{green}\frac{93}{4}\zeta(5)$$

95.$$\int_{0}^{\pi/2}\frac{\ln^2(\cos x)}{\sin^2(x)}=\color{red}\pi\color{green}\ln 2$$

96.$$\int_{0}^{\pi/2}\ln(\sin x)\ln(\cos x)\frac{\mathrm dx}{\sin^2 x}=-\color{red}\pi\left(\frac{1}{2}-\color{green}\ln 2\right)$$

97.$$\int_{0}^{\pi/4}\ln(\sin x)\ln(\cos x)\frac{\mathrm dx}{\sin^2(2x)}=\color{purple}\frac{\pi}{8}\left(\color{orange}2\ln 2-1\right)$$

98.$$\int_{0}^{\pi/4}\cot(2x)\ln(\sin x)\ln(\cos x)\frac{\mathrm dx}{\sin(2x)}=-\color{black}\frac{\ln 2}{8}\left(\color{blue}\ln 2-2\right)$$

99.$$\int_{0}^{\pi/4}\arctan(\cot^2 x)\mathrm dx=\frac{\color{red}2\pi^2-\color{blue}\ln^2(3+2\color{orange}\sqrt{2})}{16}$$

100.$$\int_{0}^{\infty}\sin(ax)\sin(bx)\frac{\ln x}{x}\mathrm dx=-\frac{1}{4}\color{red}\ln\left(\frac{b+a}{b-a}\right)\left[\color{green}2\gamma+\color{purple}\ln(b^2-a^2)\right]$$

101.$$\int_{0}^{\infty}\cos(ax)\sin(bx)\frac{\ln x}{x}\mathrm dx=-\color{red}\frac{\pi}{4}\left[2\gamma+\ln(b^2-a^2)\right]$$

101.$$ab$$

1. $$\int_{0}^{\infty}\left(\frac{1}{\sqrt{1+4x^{2\alpha}}}-e^{-x^{\alpha}}\right)\frac{\ln x}{x}\mathrm dx=-\frac{1}{2}\cdot\color{green}\left(\frac{\gamma}{\alpha}\right)^2$$

2. $$\int_{0}^{1}\frac{\ln(x)\ln^k(1-x)}{1-x}\mathrm dx=\color{blue}\Gamma(k+1)\zeta(k+2)$$

3. $$\int_{0}^{1}\frac{\ln^k(x)\ln(1-x)}{x}\mathrm dx=\color{green}\Gamma(k+1)\zeta(k+2)$$

4. $$\int_{0}^{1}\frac{\ln^k(x)\ln(1-x^2)}{x}\mathrm dx=\color{red}\left(-\frac{1}{2}\right)^{k+1}\Gamma(k+1)\zeta(k+2)$$

5. $$\int_{0}^{1}\left[\ln\left(\frac{1+x}{1-x}\right)-2\arccot(x)\right]^2\mathrm dx=\color{grey}2\zeta(2)$$

6. $$\int_{0}^{\pi/2}\frac{\ln^2(\tan^2 x)}{[\pi^2+\ln^2(\tan^2 x)]^2}\cos^2(x)\mathrm dx=\color{blue}\frac{1}{8\pi}\left[\ln 2-\frac{1}{4}\zeta(2)\right]$$

7. $$\int_{0}^{1}\left[\ln(x)\ln\left(\frac{1+x}{1-x}\right)-2\arccot(x)\right]^2\mathrm dx=\color{purple}6+2\sqrt{2}$$

8. $$\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}\right)^2\ln x \mathrm dx=\color{red}\gamma-1$$

9. $$\int_{0}^{1}\left(\frac{2-x^n}{\ln x}+\frac{1}{1-x}\right)^2\ln x \mathrm dx=\color{blue}2H_n+\color{red}\gamma-1+\color{grey}\ln\left(\frac{2n+1}{(n+1)^4}\right)$$

10. $$\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}+\frac{\ln x}{1+x}\right)^2\ln x \mathrm dx=\color{red}\gamma-1-\color{orange}\zeta(2)-\color{green}\zeta(3)$$

11. $$\int_{0}^{1}\frac{\arctan(2x)}{\sqrt{x(1-x)}}\mathrm dx=\color{red}\pi\arctan\sqrt{\phi}$$

12. $$\int_{0}^{1}\frac{x\arctan(2x)}{\sqrt{1-x^2}}\mathrm dx=\color{blue}\frac{\pi}{2\phi}$$

13. $$\int_{0}^{1}\frac{\ln\left(\frac{1}{x}-1\right)}{\sqrt{1-x^2}}\mathrm dx=-\color{red}2C$$

14. $$\int_{0}^{1}\frac{\ln\left(1-\frac{1}{x}\right)}{\sqrt{x^2-x}}\mathrm dx=\color{brown}\pi^2$$

15. $$\int_{0}^{\infty}\ln(x)\ln\left(\frac{1+x}{x}\right)\ln\left(\frac{1+x}{x^2}\right)\frac{\mathrm dx}{1+x^2}=\color{blue}\frac{\pi^2}{24}\left(16C+\color{green}\pi\ln 8\right)+\color{grey}4\beta(2)$$

16. $$\int_{0}^{\infty}\ln^2(x)\ln\left(1+\frac{1}{x^2}\right)\frac{\mathrm dx}{1-x^2}=\color{blue}\frac{\pi^4}{8}$$

17. $$\int_{0}^{\infty}\ln^4(x)\ln\left(1+\frac{1}{x^2}\right)\frac{\mathrm dx}{1-x^2}=\color{purple}\frac{\pi^6}{4}$$

18. $$\int_{0}^{\infty}\ln(x)\ln\left(1+\frac{1}{x^2}\right)\frac{\mathrm dx}{1+x}=-\color{purple}\frac{1}{16}\left(\color{orange}16\pi C-\color{blue}3\zeta(3)+\color{green}2\pi^2\ln 2\right)$$

19. $$\int_{0}^{1}\ln x\ln\left(1+\frac{1}{\sqrt{x}}\right)\mathrm dx=\color{blue}\zeta(2)-\color{green}3$$

20. $$\int_{0}^{\infty}\frac{\sqrt{1+2x^2}-x\sqrt{2}}{\sqrt{1+x^2}}\mathrm dx=\color{blue}\sqrt{2}-\frac{\color{red}\Gamma\left(\frac{2}{4}\right)\cdot\Gamma\left(\color{purple}\frac{3}{4}\right)}{\color{green}\Gamma\left(\frac{1}{4}\right)}$$

21. $$\int_{0}^{\infty}x \operatorname{csch}\left(\frac{\pi}{2}x\right)\frac{\mathrm dx}{(1+x^2)^2}=\color{green}C-\frac{1}{2}$$

22. $$\int_{0}^{\infty}e^{-x}\ln\left(e^{2x}-\sqrt{e^{2x}-1}\right)\mathrm dx=\color{green}\frac{\pi}{2}$$

23. $$\int_{0}^{\infty}e^{-2x}\ln\left(e^{x}-\sqrt{e^{2x}-1}\right)\mathrm dx=-\color{blue}\frac{1}{2}$$

24. $$\int_{0}^{\pi/2}\left(\frac{\alpha}{\ln\tan x}+\frac{1+\alpha-\tan x}{1-\tan x}\right)\mathrm dx=\color{blue}\frac{\alpha+2}{4}\cdot \color{red}\pi$$

25. $$\int_{0}^{\pi/2}\left(\frac{\alpha}{\ln\tan x}+\frac{1+\alpha-\sqrt{\tan x}}{1-\tan x}\right)\mathrm dx=\color{green}\frac{\alpha+1}{4}\cdot \color{red}\pi$$

26. $$\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{2-\sqrt[3]{\tan^4 x}}{1-\tan x}\right)\mathrm dx=\color{red}\pi$$

27. $$\int_{0}^{\pi/2}\left(\frac{\alpha}{\ln\tan x}+\frac{1+\alpha-\sqrt[n]{\tan x}}{1-\tan x}\right)\mathrm dx=\color{green}F(\alpha,n)\cdot \color{red}\pi$$

28. $$\int_{0}^{\pi/4}\left(\frac{2}{\ln\tan x}+\frac{1+\tan x}{1-\tan x}\right)\mathrm dx=\color{red}\gamma+\color{purple}2\ln \pi+\color{green}7\ln\sqrt{2}-\color{grey}4\ln\Gamma\left(\frac{1}{4}\right)$$

29. $$\int_{0}^{\pi/4}\left(\frac{1}{\ln\tan x}+\frac{\tan x}{1-\tan x}\right)\mathrm dx=\color{red}\frac{\gamma}{2}+\color{purple}\ln \pi+\color{green}\frac{7}{4}\ln 2-\color{grey}2\ln\Gamma\left(\frac{1}{4}\right)-\frac{\pi}{8}$$

30. $$\int_{0}^{\pi/4}\left(\frac{\alpha}{\ln\tan x}+\frac{1+\alpha-\tan x}{1-\tan x}\right)\mathrm dx=\color{blue}\frac{\alpha+2}{8}\cdot \color{red}\pi+\alpha\left(\color{brown}\frac{7}{4}\ln 2+\color{orange}\frac{\gamma}{2}+\ln \pi-\color{red}2\ln\Gamma\left(\frac{1}{4}\right)\right)$$

31. $$\int_{0}^{1}\ln x\ln\left(\frac{\ln x}{\ln x-1}\right)\mathrm dx=\color{grey}\gamma$$

32. $$\int_{0}^{1}\ln\left(\ln\left(\frac{1+x}{1-x}\right)\right)\frac{\mathrm dx}{(1+x)^2}=-\frac{1}{2}\color{purple}\gamma$$

33. $$\int_{0}^{1}\ln\left(\frac{1+x}{1-x}\right)\frac{\mathrm dx}{(1+x)^2}=\color{green}1$$

34. $$\int_{0}^{1}\ln\left(1+\ln\left(\frac{1+x}{1-x}\right)\right)\frac{\mathrm dx}{(1+x)^2}=\color{blue}\frac{1}{2}eE(-1)$$

35. $$\int_{0}^{\infty}x\ln(e^x-1)\frac{\mathrm dx}{e^x-1}=\color{orange}\zeta(3)$$

36. $$\int_{-\infty}^{\infty}x\sin(x)\frac{\mathrm dx}{\cos(x)+\cosh^2(x)}=2\color{blue}C-\color{red}1$$

37. $$\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}\right)\left(\frac{\alpha-x}{\ln x}+\frac{\alpha-1-\ln x}{1-x}\right)\ln(x)\mathrm dx=\color{blue}\gamma(\alpha-1)-\color{green}\alpha+2-\color{purple}\zeta(2)-\color{black}\ln 2$$

38. $$\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}\right)^2\left(\frac{\alpha-x}{\ln x}+\frac{\alpha-1-\ln x}{1-x}\right)\ln^2(x)\mathrm dx=\frac{7-3\alpha}{2}+\color{blue}\gamma(\alpha-1)+\color{purple}3\zeta(3)+(\alpha-2)\zeta(2)-\color{black}\ln 2$$

39. $$\int_{0}^{\pi/3}\sqrt{\sin^3(x)\sin\left(\frac{\pi}{3}-x\right)}\mathrm dx=\color{red}\frac{\pi}{16}$$

40. $$\int_{-\infty}^{+\infty}\frac{x^3\sin(x)}{\cos(x)+\cosh(x)}\frac{\mathrm dx}{x^2-\pi^2}=\color{blue}\pi^2\left(\color{green}\frac{\pi}{2}-\color{black}H_3\right)$$

41.$$\int_{0}^{1}\left(\frac{x}{\ln x}+\frac{1+\ln x}{1-x}\right)^2\ln(x)\mathrm dx=\color{green}\gamma-1-2\zeta(2)-\color{red}6\zeta(3)+\ln(3)$$

42. $$\int_{0}^{\pi/2}\cos(2x)\ln^2\left(\tan\left(\frac{x}{2}\right)\right)\mathrm dx=\color{green}\pi$$

43. $$\int_{0}^{\pi/2}\cos(2x)\ln^3\left(\tan\left(\frac{x}{2}\right)\right)\mathrm dx=-\color{blue}12C$$

44. $$\int_{0}^{\pi/2}\sin(2x)\ln^2\left(\tan\left(\frac{x}{2}\right)\right)\mathrm dx=\color{orange}2\ln 2$$

45. $$\int_{0}^{\pi/2}\sin(2x)\ln^3\left(\tan\left(\frac{x}{2}\right)\right)\mathrm dx=-\color{blue}\left(\frac{\pi}{2}\right)^2$$

46. $$\int_{0}^{\pi/2}\sin(x)\sin(2x)\ln^3\left(\tan\left(\frac{x}{2}\right)\right)\mathrm dx=\pi \color{red}\left(1-\color{blue}\frac{\pi^2}{8}\right)$$

47. $$\int_{0}^{\pi/2}\cos(x)\cos(2x)\ln^3\left(\tan\left(\frac{x}{2}\right)\right)\mathrm dx=-\pi \color{red}\left(1+\color{green}\frac{\pi^2}{4}\right)$$

48. $$\int_{0}^{\pi/2}\cos(x)\cos(2x)\ln^2\left(\tan\left(\frac{x}{2}\right)\right)\mathrm dx=\frac{\color{red}2}{\color{blue}3}\color{red}\left(\color{green}4C+\color{gre}1\right)$$

49. $$\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{\mathrm dx}{1+x^3}=-\color{green}\frac{2\pi}{27}\left[\color{red}\pi+\color{blue}(1-3\ln 3)\sqrt{3}\right]$$

50. $$\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{x^3}{1+x^3}\mathrm dx=\color{green}\frac{\pi}{27}\left[\color{blue}(2+3\ln 3)\sqrt{3}-\pi\right]$$

51. $$\int_{0}^{\infty}\frac{\ln(1+x^3)}{\sqrt{1+x^3}}\mathrm dx=\color{red}\frac{\Gamma^3\left(\frac{1}{3}\right)}{2^{\frac{4}{3}}}\left(\color{green}1+\color{purple}\sqrt{3}\cdot\color{brown}\frac{\ln 3}{\pi}\right)$$

52.$$\int_{0}^{\infty}\prod_{k=1}^{2n-1}\left(1+\frac{x^2}{k^2}\right)^{-1}\mathrm dx=\color{blue}\frac{\pi}{2}\cdot\frac{2n-1}{4n-3}$$

53.$$\int_{0}^{\infty}\prod_{k=1}^{2n}\left(1+\frac{x^2}{k^2}\right)^{-1}\mathrm dx=\color{red}\frac{\pi}{2}\cdot\frac{2n}{4n-3}$$

54.$$\int_{0}^{\pi/2}\sin^2\left(\frac{x}{2}\right)\frac{\ln \sin x}{\sin x}\mathrm dx=\color{blue}\frac{\eta^2(1)-\eta(2)}{4}$$

55.$$\int_{0}^{\pi/2}\sin^2\left(\frac{x}{2}\right)\frac{\ln \sin x}{\sin^2 x}\mathrm dx=\color{brown}\frac{2-\pi}{4}$$

56.$$\int_{0}^{\pi/2}\sin^2\left(\frac{x}{2}\right)\frac{\ln^2 \sin x}{\sin^2 x}\mathrm dx=\color{green}\frac{2-\pi(1-\ln 2)}{2}$$

57.$$\int_{0}^{\pi/2}\frac{\sin^2\left(\frac{x}{2}\right)}{\sin^2(x)}\ln^k\tan\left(\frac{x}{2}\right)\mathrm dx=\color{green}\frac{(-1)^k}{2}\Gamma(1+k)$$

58.$$\int_{0}^{\pi/2}\ln\sin x \ln\tan\left(\frac{x}{2}\right)\frac{\sin^2\left(\frac{x}{2}\right)}{\sin^2 x}\mathrm dx=\color{green}C+\color{red}\beta(1)-\color{purple}1$$

59.$$\int_{0}^{\pi/2}\ln\left(\sin x\cos\left(\frac{x}{2}\right)\right) \ln\tan\left(\frac{x}{2}\right)\frac{\sin^2\left(\frac{x}{2}\right)}{\sin^2 x}\mathrm dx=\frac{12\color{red}C+3\color{blue}\pi+\color{green}\ln 4-16}{8}$$

60.$$\int_{0}^{\pi/2}\ln\left(\sin\left(\frac{x}{2}\right)\right)\ln\left(\cos\left(\frac{x}{2}\right)\right)\frac{\sin^2\left(\frac{x}{2}\right)}{\sin^2 x}\mathrm dx=\frac{2\color{red}(\pi-1)\color{purple}\eta(1)+\color{green}\eta^2(1)-\color{orange}\pi}{8}$$

61.$$\int_{0}^{\pi/2}\ln\left(\sin\left(\frac{x}{2}\right)\right)\ln\left(\cos\left(\frac{x}{2}\right)\right)\frac{\mathrm dx}{\sin^2(x)}=-\color{green}\beta(1)[1-2\color{yellow}\eta(1)]$$

62.$$\int_{0}^{\pi/2}\cot(x)\ln\left(\sin\left(\frac{x}{2}\right)\right)\ln\left(\cos\left(\frac{x}{2}\right)\right)\frac{\mathrm dx}{\sin(x)}=\frac{1}{4}[2\color{red}\eta(1)-\color{blue}\eta^2(1)]$$

63.$$\int_{0}^{\pi/2}\ln\left(\sin\left(\frac{x}{2}\right)\right)\ln\left(\cos\left(\frac{x}{2}\right)\right)\frac{\mathrm dx}{1+\cos(x)}=\color{grey}\frac{\eta(1)}{2}\left(\frac{\eta(1)}{2}+\color{blue}\pi-1\right)-\color{green}\beta(1)$$

64.$$\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{\tan x}{1-\tan x}\right)^2\mathrm dx=\color{blue}3\ln(2)-4\color{grey}\frac{C}{\pi}-\color{brown}\frac{1}{2}$$

65.$$\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{\tan x}{1-\tan x}\right)^3\mathrm dx=\color{blue}\frac{9}{2}\ln(2)-6\color{grey}\frac{C}{\pi}-\color{brown}\frac{3}{4}-\color{black}\frac{\pi}{8}$$

66.$$\int_{0}^{2\pi}\frac{[a+b\sin^c(x)]^d}{[a+b\sin^c(x)]^d+[a+b\cos^c(x)]^d}=\color{grey}\pi$$

67.$$\int_{0}^{\pi/2}x\cot(x)\arctan\left(\cot(x)\right)\mathrm dx=\frac{7}{8}\color{blue}\zeta(3)$$

68.$$\int_{0}^{\pi/2}x^2\cot(x)\arctan\left(\cot(x)\right)\mathrm dx=\frac{1}{8}\color{blue}\pi\color{green}\zeta(3)$$

69.$$\int_{0}^{\pi/2}x^3\cot(x)\arctan\left(\cot(x)\right)\mathrm dx=\color{grey}\frac{1}{32}\left[9\pi^2\zeta(3)-\color{purple}93\zeta(5)\right]$$

70.$$\int_{0}^{\pi/2}x^4\cot(x)\arctan\left(\cot(x)\right)\mathrm dx=\frac{3}{16}\pi\left[9\color{red}\pi^2\zeta(3)-\color{green}11\zeta(5)\right]$$

71.$$\int_{0}^{\pi/2}x^5\cot(x)\arctan\left(\cot(x)\right)\mathrm dx=\frac{1}{256}\left[30\pi^4\color{red}\zeta(3)-900\color{green}\pi^2\color{brown}\zeta(5)+\color{black}5715\zeta(7)\right]$$

72.$$\int_{0}^{\pi/2}x^6\cot(x)\arctan\left(\cot(x)\right)\mathrm dx=\color{green}\frac{1}{256}\pi\left[18\pi^4\color{red}\zeta(3)-900\color{green}\pi^2\color{brown}\zeta(5)+\color{black}7065\color{yellow}\zeta(7)\right]$$

73.$$\int_{0}^{\pi/2}\arctan\left(\frac{\sin x}{\cos x-1-\sqrt{2}}\right)\frac{\mathrm dx}{\tan x}=\color{green}\frac{\pi\color{blue}\ln(2\sqrt{2}-2)-2\color{red}C}{\color{grey}4}$$

74.$$\int_{0}^{\infty}\frac{\cos(10\pi x^2)\sin(6\pi x^2)}{\sinh^2(2\pi x)}\mathrm dx=\color{blue}\frac{1}{16}$$

75.$$\int_{0}^{\infty}xe^{-x}\sin x\frac{x^2+3x+3}{(1+x)^3}\mathrm dx=\int_{0}^{\infty}e^{-x}\cos x\frac{(x^2+3x+3)^2}{1+x}\mathrm dx$$

76.$$\int_{0}^{\infty}x\sin^2(x)\frac{\mathrm dx}{\sinh(x)}=\color{blue}\frac{\pi^2}{8}\color{green}\tanh^2(\pi)$$

77.$$\int_{0}^{\infty}x\sin^4(x)\frac{\mathrm dx}{\sinh(x)}=\color{red}\frac{\pi^2}{8}\color{purple}\frac{2+\tanh^2(\pi)}{[1+\coth^2(\pi)]^2}$$

78.$$\int_{0}^{\infty}x\cos(2x)\frac{\mathrm dx}{\sinh(x)}=\color{blue}\frac{\pi^2}{4}\color{green}\operatorname{sech^2(\pi)}$$

79.$$\int_{-\infty}^{+\infty}x\cos^2\left(x/2\right)\frac{\mathrm dx}{\sinh(x)}=\color{blue}\left(\frac{\pi}{2}\right)^2\color{green}\cdot\frac{3+\color{red}\cosh(\pi)}{1+\color{black}\cosh(\pi)}$$

80. $$\int_{0}^{\infty}\operatorname{arctanh(e^{-x})}\mathrm dx=\color{red}\lambda(2)$$

81. $$\int_{0}^{\infty}\operatorname{arctanh^2(e^{-x})}\mathrm dx=\int_{0}^{\infty}x\operatorname{arctanh(e^{-x})}\mathrm dx=\color{orange}\lambda(3)$$

82.$$\int_{0}^{\infty}\sqrt[4]{e^x-1}\cdot\frac{\mathrm dx}{1-2\cosh x}=\frac{2\pi}{3+\sqrt{3}}$$

83.$$\int_{0}^{\infty}\sqrt[3]{e^x-1}\cdot\frac{\mathrm dx}{1-2\cosh x}=-\color{blue}\frac{2\pi}{3}\left[\sin\left(\frac{\pi}{9}\right)-\sin\left(\frac{2\pi}{9}\right)+\cos\left(\frac{\pi}{18}\right)\right]$$

84.$$\int_{0}^{\infty}\sqrt[6]{e^x-1}\cdot\frac{\mathrm dx}{1-2\cosh x}=\color{grey}\frac{2\pi}{3}\left[\sin\left(\frac{\pi}{9}\right)-3\sin\left(\frac{2\pi}{9}\right)+\cos\left(\frac{\pi}{18}\right)\right]$$

85.$$\int_{0}^{\infty}\sqrt[3]{(e^x-1)^2}\cdot\frac{\mathrm dx}{1-2\cosh x}=-\frac{4}{3}\color{red}\pi\color{blue}\sin\left(\frac{2\pi}{9}\right)$$

86.$$\int_{0}^{\pi/2}\arctan[2\alpha^2\tan^2(x)]\mathrm dx=\color{red}\frac{\pi}{2}\color{blue}\arctan\left(\frac{2\alpha^2+2\alpha}{2\alpha+1}\right)$$

87.$$\int_{0}^{\pi/2}\ln[x^2+\ln^2(3\cos x)]\mathrm dx=\color{blue}\pi\color{red}\ln\ln\left(\frac{27}{8}\right)$$

88.$$\int_{0}^{\pi/2}\ln[x^2+\ln^2(4\cos x)]\mathrm dx=\color{blue}\pi\color{purple}\ln\ln 4$$

89.$$\int_{0}^{\pi/2}\ln[x^2+\ln^2(6\cos x)]\mathrm dx=\color{blue}\pi\color{green}\ln\ln\sqrt{27}$$

90.$$\int_{0}^{\pi/2}\ln\left[x^2+\ln^2\left(\frac{1}{\alpha}\cos x\right)\right]\mathrm dx=\color{blue}\pi\color{grey}\ln\ln(2\color{purple}\alpha)$$

91.$$\int_{0}^{\pi/2}\cos^{2n}(x)\ln\sin(x)\mathrm dx=-\color{red}\frac{(2n-1)!!}{(2n)!!}\left(\color{green}\frac{H_n}{2}+\ln 2\right)\color{blue}\pi$$

92.$$\int_{0}^{\pi}\cos(x)\cos[(2n+1)x]\ln\sin(x)\mathrm dx=-\color{blue}\frac{2n+1}{n(n+1)}\color{red}\cdot\frac{\pi}{4}$$

93.$$\int_{0}^{\pi/2}\sin^k(x)\cos(x)\ln^n\sin(x)\mathrm dx=\color{red}(-1)^n\color{green}\frac{n!}{(k+1)^{n+1}}$$

94.$$\int_{0}^{1}\frac{\ln x}{x}\cdot \frac{\ln(1+x)}{1+x}\mathrm dx=-\frac{5}{8}\color{blue}\zeta(3)$$

95.$$\int_{0}^{1}\frac{\ln x}{x}\cdot \frac{\ln(1-x)}{1-x}\mathrm dx=2\color{green}\zeta(3)$$

96.$$\int_{0}^{\pi/4}\left(\frac{\ln\sin(x)}{\cot(x)}+\frac{\ln\cos(x)}{\tan(x)}\right)\sin^{2n+1}(x)\mathrm dx=\color{red}\frac{\pi}{2}\cdot\color{blue}\frac{(2n-1)!!}{(2n)!!}\left(\color{green}\frac{H^*_{2n+1}}{2}-\color{brown}\ln 2\right)$$

97.$$\int_{0}^{\pi/2}\tan(x)\ln\cos(x)\ln\sin^2(x)=-\color{red}\frac{\zeta(4)}{2^5}$$

98.$$\int_{0}^{\pi/2}\cot(x)\ln\cos^2(x)\ln\sin(x)=-\color{green}\frac{\zeta(4)}{2^5}$$

99.$$\int_{0}^{\infty}\frac{\cos\left(2^{2n+1}\pi x^2\right)}{\cosh^2(2^n \pi x)}\mathrm dx=\color{green}\frac{1}{2^{n+2}}$$

100.$$\int_{0}^{\infty}\frac{\sin\left(2^{2n+1}\pi x^2\right)}{\cosh^2(2^n \pi x)}\mathrm dx=\color{green}\frac{1}{2^{n+2}}-\color{blue}\frac{1}{2^{n+1}\pi}$$

1. $$\int_{0}^{\pi/4}\ln\left(1+\sqrt[2n]{\tan x}\right)\frac{\mathrm dx}{\cos^2(x)}=\color{red}H_{2n}^{'}$$

2. $$\int_{0}^{\pi/4}\ln\left(1-\sqrt[n]{\tan x}\right)\frac{\mathrm dx}{\cos^2(x)}=-\color{red}H_n$$

3. $$\int_{0}^{\pi/4}\ln\tan\left(\frac{x}{2}\right)\frac{\mathrm dx}{\cos^2(x)}=\color{brown}\ln\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)$$

4. $$\int_{0}^{\pi/4}\ln\left(\frac{1-\sqrt[n]{\tan x}}{1-\sqrt[n+1]{\tan x}}\right)\frac{\mathrm dx}{\cos^2(x)}=\color{grey}\frac{1}{n}$$

5. $$\int_{0}^{\pi/4}\ln\left[1-\tan\left(x-\frac{\pi}{4}\right)\right]\frac{\mathrm dx}{\cos^2(x)}=-\color{red}\ln 2$$

6. $$\int_{0}^{\pi/4}\ln\left[1+\tan(x)\tan(2x)\right]\frac{\mathrm dx}{\cos^2(x)}=\color{blue}\frac{\pi}{2}-\ln 2$$

7. $$\int_{0}^{\pi/4}\ln\left[2+\tan(x)\tan(2x)\right]\frac{\mathrm dx}{\cos^2(x)}=\color{green}{2}-\ln 2$$

8. $$\int_{0}^{\pi/4}\frac{\ln[\tan x \tan(2x)]}{\cos^2(x)}\mathrm dx=-\color{orange}\ln 2$$

8. $$\int_{0}^{\pi/4}\frac{\ln^2(1+\tan x)}{\cos^2(x)}\mathrm dx=\color{purple}2(\ln 2-1)^2$$

9. $$\int_{0}^{\pi/4}\frac{(2+\ln\tan x)\ln(\tan x)\ln(1+\tan x)}{\cos^2(x)}\mathrm dx=-\color{red}\frac{3}{2}\zeta(3)-\color{blue}2$$

10. $$\int_{0}^{\pi/4}\frac{(2^2+\ln^2\tan x)\ln(\tan x)\ln(1+\tan x)}{\cos^2(x)}\mathrm dx=-\frac{1}{2}\color{grey}\left[\frac{21}{2}\zeta(4)+9\zeta(3)+10\zeta(2)+40\ln 2-64\right]$$

11. $$\int_{0}^{\pi/4}\frac{\ln^3(\tan x)\ln(1+\tan x)}{\cos^2(x)}\mathrm dx= \color{blue}-\frac{21}{4}\zeta(4)-\frac{9}{2}\zeta(3)-3\zeta(2)-12\ln 2+24$$

12. $$\int_{0}^{\pi/4}\frac{\ln^2(\tan x)\ln(1+\tan x)}{\cos^2(x)}\mathrm dx= \color{green}\frac{3}{2}\zeta(3)+\zeta(2)+4\ln 2-6$$

13. $$\int_{0}^{\pi/4}\frac{\ln(\tan x)\ln(1+\tan x)}{\cos^2(x)}\mathrm dx= \color{brown}-\frac{1}{2}\zeta(2)-2\ln 2+2$$

14. $$\int_{0}^{\pi/4}\frac{\ln(\tan x)\ln^2(1+\tan x)}{\cos^2(x)}\mathrm dx= \color{green}-\frac{1}{4}\zeta(3)+\zeta(2)-2\ln^2(2)+8\ln 2-6$$

15. $$\int_{0}^{\infty}\frac{\ln(1+x^{-4})}{1+x^4}\mathrm dx=\color{brown}\frac{1}{2}\pi \ln 2$$

17. $$\int_{0}^{\infty}\frac{\ln(1+x^{-2})}{1+x^2}\mathrm dx=\color{purple}\pi \ln 2$$

18. $$\int_{0}^{\infty}\frac{\ln(1+x^{-2})}{\sqrt{1+x^2}}\mathrm dx=\color{orange}\left(\frac{\pi}{2}\right)^2$$

19. $$\int_{0}^{\infty}\ln x\frac{\ln(1+x^{-1})}{1+x}\mathrm dx=\color{purple}-\zeta(3)$$

20. $$\int_{0}^{\infty}\ln x \ln(1+x)\frac{\ln(1+x^{-1})}{1+x}\mathrm dx=\color{green}\frac{3}{2}\zeta(4)$$

21. $$\int_{0}^{\infty}\ln x \ln(2+x)\frac{\ln(1+x^{-1})}{1+x}\mathrm dx=\color{blue}\frac{7}{16}\zeta(4)$$

22. $$\int_{0}^{\infty}\ln x\ln(1+x)\frac{\ln^2(1+x^{-1})}{1+x}\mathrm dx=\color{green}2\zeta(2)\zeta(3)-\color{red}4\zeta(5)$$

23. $$\int_{0}^{\infty}\ln^2(x)\ln(1+x)\frac{\ln^2(1+x^{-1})}{1+x}\mathrm dx=\color{brown}6\zeta^2(3)-\color{purple}\frac{15}{2}\zeta(6)$$

24. $$\int_{0}^{\infty}\ln^2(x)\ln(1+x)\frac{\ln^2(1+x^{-1})}{x}\mathrm dx=\color{brown}6\zeta^2(3)-\color{purple}\frac{63}{4}\zeta(6)$$

25. $$\int_{0}^{\infty}\ln x\ln(1+x)\frac{\ln^2(1+x^{-1})}{x}\mathrm dx=-\frac{\color{green}\zeta(2)\color{blue}\zeta(3)+\color{red}\zeta(5)}{2}$$

26. $$-(2n+1)\int_{0}^{\infty}(1-x\arccot x)\ln^{2n}(x)\mathrm dx=2\int_{0}^{\infty}(1-x\arccot x)\ln^{2n+1}(x)\mathrm dx$$

(26): $$n\ge 2$$

27. $$\int_{0}^{\infty}x\arccot x(1-x\arccot x)(2-x\arccot x)\mathrm dx=\frac{1}{4}\color{blue}\left(\frac{\pi}{2}-1\right)\color{red}\left(\frac{\pi}{2}+0\right)\color{green}\left(\frac{\pi}{2}+1\right)$$

28. $$\int_{0}^{\infty}x\arccot x(1-x\arccot x)(2+x\arccot x)\mathrm dx=\color{blue}\frac{2^5}{5^2}$$

29. $$\int_{0}^{\pi/3}\arccos\sqrt{\frac{1+\cos x}{1+2\cos x}}\mathrm dx=\color{red}\left(\frac{\pi}{4}\right)^2$$

30. $$\int_{0}^{\pi/2}\arccos\sqrt{\frac{1+\cos x}{1+2\cos x}}\mathrm dx=\color{green}\frac{\zeta(2)}{2}$$

31. $$\int_{0}^{\infty}\frac{\ln(\alpha+\arctan x)}{1+x^2}\mathrm dx=\color{blue}\frac{\pi}{2}\left[\ln\left(\frac{2\alpha+\pi}{2}\right)-1\right]+\color{green}\alpha\ln\left(\frac{2\alpha+\pi}{2\alpha}\right)$$

32. $$\int_{0}^{\infty}\frac{\ln(\alpha+\arctan^2 x)}{1+x^2}\mathrm dx=\color{red}\frac{\pi}{2}\left[\ln\left(\frac{4\alpha+\pi^2}{4}\right)-2\right]+\color{green}2\sqrt{\alpha}\arctan\left(\frac{\pi}{2\sqrt{\alpha}}\right)$$

33. $$\int_{0}^{\infty}\frac{\ln\left[\left(\frac{\pi}{2}\right)^2-\arctan^2(x)\right]}{1+x^2}\mathrm dx=\color{brown}\pi\ln\left(\frac{\pi}{e}\right)$$

34. $$\int_{0}^{\infty}\frac{\ln\left(\frac{\pi}{2}-\arctan x\right)}{1+x^2}\mathrm dx=\int_{0}^{\infty}\frac{\ln\left(\arctan x\right)}{1+x^2}\mathrm dx=\color{grey}\frac{\pi}{2}\left(\ln\frac{\pi}{2}-1\right)$$

35. $$\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}\ln\left(\frac\right)\mathrm dx=\color{blue}\left(\frac{\pi}{2}\right)^2\ln(2)$$

36. $$\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}\ln\left(\frac\right)\mathrm dx=\color{green}\left(\frac{\pi}{4}\right)^2\color{purple}\ln\left(\frac{\pi}{4}\right)$$

37. $$\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}\ln(\arctan(x))\mathrm dx=-\color{brown}\left(\frac{\pi}{4}\right)^2\left[\ln\left(\frac{2}{\pi}\right)^2+1\right]$$

38. $$\int_{0}^{\infty}\frac{1-x}{1+x}\arctan(x)\ln\left(\frac{1+x^2}{2}\right)\mathrm dx=\color{red}\left(\frac{\pi}{2}\right)^3\cdot \color{green}\frac{1}{3!}$$

39. $$\int_{0}^{\pi/2}\frac{\arctan(2\cos^2 x)}{\cos^2 x}\mathrm dx=\color{purple}\frac{\pi}{\sqrt{\phi}}$$

40. $$\int_{0}^{\pi/2}\frac{\arctan(2\sec x)}{\sec x}\mathrm dx=\color{blue}\frac{\pi}{\phi^2}$$

41. $$\int_{0}^{\pi/2}\frac{\arctan\left(\frac{1}{2}\sec x\right)}{\sec x}\mathrm dx=\color{green}\frac{\pi}{2\phi^2}$$

42. $$\int_{-1}^{+1}\frac{x}{1-x^2}\arctan\left(\frac{\sqrt{1-x^2}}{x}\right)\mathrm dx=\color{green}\pi \ln 2$$

43. $$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{red}\frac{\zeta(2)}{2}$$

44. $$\int_{0}^{1}\int_{0}^{1}\frac{1}{2}\cdot\frac{\ln^2(x)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{orange}\frac{\zeta(3)}{3}$$

45. $$\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)\ln(y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{purple}\frac{\zeta(3)}{3}$$

46. $$\int_{0}^{1}\int_{0}^{1}\frac{\sqrt{x}-1}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{bown}\gamma+\ln\frac{\pi}{4}$$

47. $$\int_{0}^{1}\int_{0}^{1}\frac{x(y-1)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{black}1-\gamma$$

48. $$\int_{0}^{1}\int_{0}^{1}\frac{x(y^2-1)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{green}\ln(2)$$

49. $$\int_{0}^{1}\int_{0}^{1}\frac{x^2(y^2-1)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{green}\frac{1}{2}\ln(2)$$

50. $$\int_{0}^{1}\int_{0}^{1}\frac{1}{{2}{\phi^{-1}}-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=-\color{blue}\ln(2\phi^2)$$

51. $$\int_{0}^{1}\int_{0}^{1}\frac{\sqrt{x}-x}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{green}\ln\frac{\pi}{4}$$

52. $$\int_{0}^{1}\int_{0}^{1}\frac{\sqrt{x}-x^2}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{blue}\ln\left(\frac{\pi}{4}\cdot\frac{\sqrt{2}}{2}\right)$$

53. $$\int_{0}^{1}\int_{0}^{1}\frac{\ln(xy)}{1-xy}x{\mathrm dx \mathrm dy}=-\color{purple}1$$

54. $$\int_{0}^{1}\int_{0}^{1}\frac{\ln(xy)}{1+xy}x{\mathrm dx \mathrm dy}=\color{brown}1-\zeta(2)$$

55. $$\int_{0}^{1}\int_{0}^{1}\frac{\ln(xy)}{1-xy}xy{\mathrm dx \mathrm dy}=\color{red}2[1-\zeta(2)]$$

56. $$\int_{0}^{1}\int_{0}^{1}\frac{\ln(xy)}{1+xy}xy{\mathrm dx \mathrm dy}=\color{grey}\frac{3\zeta(3)-4}{2}$$

57. $$\int_{0}^{1}\int_{0}^{1}\frac{x\ln(x)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{purple}1-\gamma$$

58. $$\int_{0}^{1}\int_{0}^{1}\frac{x\ln(x)\ln(y)}{1-xy}\cdot \frac{\mathrm dx \mathrm dy}{\ln(xy)}=\color{green}1-2\gamma$$ 58. $$\int_{0}^{1}\frac{\ln(x)\ln(1-x)\ln[x(1-x)]}{\sqrt{x(1-x)}}\mathrm dx=\color{blue}4\pi[\zeta(3)-\color{red}4\ln^3(2)]$$

59. $$\int_{0}^{\pi/4}\sin(8x)\ln^3[\cot(x)]\mathrm dx=\color{red}1+\eta(2)$$

60. $$\int_{0}^{\pi/2}\sin(8x)\ln^3[\cot(x)]\mathrm dx=\color{blue}2+\zeta(2)$$

61. $$\int_{0}^{\pi/2}\sin(2x)\ln^2[\cot(x)]\mathrm dx=\color{green}\eta(2)$$

62. $$\int_{0}^{\pi/2}x\sin(8x)\ln^2[\cot(x)]\mathrm dx=\color{grey}\zeta(2)$$

63. $$\int_{0}^{\pi/4}\cos(8x)\ln^3[\cot(x)]\mathrm dx=\color{brown}4C+1$$

63. $$\int_{0}^{\pi/4}\cos(4x)\ln^3[\cot(x)]\mathrm dx=6C$$

64. $$\int_{0}^{\pi/2}\cos(4x)\ln^2[\cot(x)]\mathrm dx=\color{yellow}\pi$$

65. $$\int_{0}^{\pi/4}\cos(4x)\ln^2[\cot(x)]\mathrm dx=\color{red}\frac{\pi}{2}$$

66. $$\int_{0}^{\pi/4}\cos(8x)\ln^2[\cot(x)]\mathrm dx=\frac{\pi}{3}$$

67. $$\int_{0}^{\pi/2}\sin(4x)\ln^3[\cot(x)]\mathrm dx=\frac{\pi^2}{4}$$

68. $$\int_{0}^{\pi/2}x\cos(4x)\ln^2[\cot(x)]\mathrm dx=\color{purple}\frac{\pi^2}{4}$$

69. $$\int_{0}^{\pi/4}x\sin(4x)\ln^3[\cot(x)]\mathrm dx=\color{brown}\frac{\pi^2}{8}$$

70. $$\int_{0}^{\pi/4}x\sin(4x)\ln^2[\cot(x)]\mathrm dx=\color{green}\ln(2)$$

71. $$\int_{0}^{\pi/4}x\cos(4x)\ln^2[\cot(x)]\mathrm dx=\color{grey}\frac{1}{16}[\pi^2-12\ln(2)]$$

72. $$\int_{0}^{\pi/4}\sin(4x)\ln\cot(x)\mathrm dx=\color{green}\frac{1}{2}$$

73. $$\int_{0}^{\pi/4}\sin(8x)\ln\cot(x)\mathrm dx=\color{blue}\frac{1}{3}$$

74. $$\int_{0}^{\pi/2}\arctan[2\sin^2(x)]\mathrm dx=\color{blue}\frac{\pi}{2}\arccot\sqrt{\phi}$$

75. $$\int_{0}^{\pi/2}\cos(2x)\arctan[2\sin^3(x)]\mathrm dx=\color{green}\frac{1}{2}\cdot [\sqrt{2}\ln(3+2\sqrt{2})-\pi]$$

76. $$\sum_{k=0}^{N}\int_{0}^{\pi/2}\sin(2x)\arctan[2k+\cos(2x)]\mathrm dx=\color{blue}\frac{-\pi-2\ln(2N^2+2N+1)+4(2N+1)\arctan(2N+1)}{8}$$

77. $$\int_{0}^{\pi/2}\sin^2(x)\arctan\left[1-\frac{1}{4}\sin^2(2x)\right]\mathrm dx=\color{green}\frac{\pi}{2}\left(\frac{\pi}{4}-\arctan\sqrt{\frac{\sqrt{2}-1}{2}}\right)$$

78. $$\int_{0}^{\infty}\frac{\arctan(ax)-\arctan(bx)}{x}\ln(x)\mathrm dx=-\color{blue}\frac{\pi}{4}\ln\left[\ln^2(a)-\ln^2(b)\right]$$

79. $$\int_{0}^{\infty}\frac{\arctan(ax)-2\arctan(bx)+\arctan(cx)}{x}\ln(x)\mathrm dx=-\color{green}\frac{\pi}{4}\ln\left[\ln^2(a)-2\ln^2(b)+\ln^2(c)\right]$$

80. $$\int_{0}^{\infty}\frac{\sin(x)\ln(x)}{x}\mathrm dx=-\color{red}\frac{\pi \gamma}{2}$$

81. $$\int_{0}^{\infty}\frac{\sin(\ln x)}{x\ln x}\mathrm dx=\color{orange}\pi$$

82. $$\int_{0}^{\infty}\frac{\arctan(x)\ln^2(x)}{1+x^2}\mathrm dx=\color{purple}3\lambda(4)$$

83. $$\int_{0}^{\pi}\int_{0}^{\pi}xy\ln[1-\cos(2x)\sin(2y)]\mathrm dx \mathrm dy=\color{grey}(\pi)^4\left(\frac{C}{\pi}-\frac{1}{2}\ln 2\right)$$

84. $$\int_{0}^{\pi}\int_{0}^{\pi}xy\ln[1-\cos^2(2x)\sin^2(2y)]\mathrm dx \mathrm dy=\color{blue}\pi^3\left(2C-\pi\ln 2\right)$$

85. $$\int_{0}^{1}\frac{1}{\arctan^2\left(\frac{1+2x-x^2}{1-2x-x^2}\right)}\frac{\mathrm dx}{1+x^2}\mathrm dx=\color{blue}\frac{2}{\pi}$$

86. $$\int_{0}^{2}\frac{1}{\arctan\left(\frac{1+2x-x^2}{1-2x-x^2}\right)}\frac{\mathrm dx}{1+x^2}\mathrm dx=\color{green}\frac{1}{2}\ln\left(\frac{4}{\pi}\arccot(7)\right)$$

87. $$\int_{0}^{2}\frac{1}{\arctan^2\left(\frac{1+2x-x^2}{1-2x-x^2}\right)}\frac{\mathrm dx}{1+x^2}\mathrm dx=\color{red}\frac{1}{2\arccot(7)}$$

88. $$\int_{0}^{\infty}\frac{\ln x}((1+x)(1+x^2)(1+x^3))\mathrm dx=-\frac{2}{3^3}\color{red}\pi^2$$

89. $$\int_{0}^{1}\arctan(x)\ln\left(\frac{1+x^2}{2x}\right)\mathrm dx=\color{blue}\frac{\pi^2}{4!}-\color{red}\frac{\pi}{4}+\left(\frac{\ln 2}{2}\right)^2+\color{green}\frac{\ln 2}{2}$$

90. $$\int_{0}^{1}\frac{\arctan(x)\ln^2(x)}{x}\mathrm dx=\color{red}\beta(2)$$

91. $$\int_{0}^{\pi/2}\sin(x)\ln[\alpha^2+\sec^2(x)]\mathrm dx=\color{red}\ln(1+\alpha^2)+\color{green}\frac{1}{\alpha}\arctan(\alpha)$$

92. $$\int_{0}^{\pi/2}\cos(x)\ln[\alpha^2+\tan^2(x)]\mathrm dx=\color{red}\sqrt{\frac{\alpha}{\alpha-1}}\color{grey}\ln\left[2\alpha-1+2\sqrt{\alpha(\alpha-1)}\right]-\color{green}\ln 4$$

93. $$\int_{0}^{\pi/2}\sin(2x)\ln[\alpha+\tan^2(x)]\mathrm dx=\color{red}\frac{\alpha}{\alpha-1}\ln \alpha$$

94. $$\int_{0}^{\pi/2}\cos(2x)\ln[\alpha+\tan^2(x)]\mathrm dx=\color{blue}\frac{\pi}{\alpha-1}\color{red}\left(1-\sqrt{\alpha}\right)$$

95. $$\int_{0}^{\pi/2}\cos(2x)\ln[2^2+\tan^2(x)]\mathrm dx=-\color{blue}\frac{1}{2}\cdot\color{green}\frac{\pi}{\phi}$$

96. $$\int_{0}^{\pi/2}\cos(4x)\ln[\alpha^2+\tan^2(x)]\mathrm dx=\color{purple}\frac{\alpha}{(\alpha+1)^2}\color{gre}\pi$$

97. $$\int_{0}^{\pi/2}\cos(4x)\ln[\phi^{-4}+\tan^2(x)]\mathrm dx=\color{brown}\frac{\pi}{5}$$

98. $$\int_{0}^{\pi/2}\cos(4x)\ln\left[2+\tan(x)\tan\left(\frac{x}{2}\right)\right]\mathrm dx=\color{blue}\frac{1}{3}$$

99. $$\int_{0}^{\pi/2}\cos(2x)\ln\left[1+\tan(x)\tan\left(\frac{x}{2}\right)\right]\mathrm dx=-\color{green}\frac{\pi}{4}$$

1. $$\int_{0}^{1}\frac{x}{1-x^2}\ln\left(\frac{1+x}{1+x^2}\right)\mathrm dx=\color{red}{\frac{1}{4}\ln^2(2)}$$

2. $$\int_{0}^{\pi/2}\arctan\left(\frac{\ln\sin x}{x}\right)\mathrm dx=\color{blue}{-\frac{\pi}{2}}\arctan\left(\frac{2}{\pi}\cdot\ln 3\right)$$

3. $$\int_{0}^{\infty}\frac{\tanh^2(x)\tanh(2x)}{x^2}\mathrm dx=\color{green}{2\ln 2}$$

4. $$\int_{0}^{1}t^2\cos(2t\pi)\tan(t\pi)\ln[\sin(t\pi)]\mathrm dt=\frac{1}{\pi}\cdot\color{green}{{\frac{\ln 2}{2}}(1-\ln 2)}$$

5. $$\int_{0}^{\infty}\frac{(x+\sqrt{5})(x-1)}{\sqrt{2^x-1}\ln(2^x-1)}\mathrm dx=\color{red}{\pi}\color{brown}{\left(\frac{\phi}{\ln 2}\right)^2}$$

6. $$\int_{0}^{\pi/6}\arctan\sqrt{2-\tan^2(x)}\mathrm dx=\color{purple}\frac{\pi}{20}$$

7. $$\int_{0}^{\infty}\frac{\ln x}{(1+x)(1+x^2)(1+x^3)}\mathrm dx=\int_{0}^{\infty}\frac{\ln x}{(1+x)(1+x^2)(1+x^3)(1+x^4)}\mathrm dx=-\color{blue}{\frac{4\pi^2}{27}}$$

8. $$\int_{0}^{\pi/2}\frac{\ln(2\cos x)}{x^2+\ln^2(2\cos x)}\mathrm dx=\color{red}\frac{\pi}{4}$$

9. $$\int_{0}^{\pi/2}\frac{\sin^{2k+1}(2x)}{1+\sqrt{\tan x}}\mathrm dx=\frac{1}{2} \cdot \color{green}\frac{(2k)!!}{(2k+1)!!}$$

10. $$\int_{0}^{\infty}x \arccot(x)\ln\left(1+\frac{1}{x^2}\right)\mathrm dx=\color{brown}\frac{\pi}{2}$$

11. $$\int_{-\pi/3}^{\pi/2}\arccos\left(\frac{1}{1+2\cos x}\right)\mathrm dx=\frac{7}{24}\cdot \pi^2$$

12. $$\int_{0}^{\pi/2}\frac{\ln [\sin^2(x)]}{\sin(2x)}\sqrt[5]{\tan x}\mathrm dx=-\color{green}{5\pi\phi}$$

13. $$\int_{0}^{\infty}\ln(x) \ln\left(1+\frac{1}{x^2}\right)=-\color{red}\pi$$

14. $$\int_{0}^{\infty}x\ln^2\left(1+\frac{1}{x^2}\right)=\zeta(2)$$

15. $$\int_{0}^{\infty}\arctan(x)\ln\left(1+\frac{1}{x^2}\right)=\color{red}\zeta(2)$$

16. $$\int_{0}^{\pi/2}\cos(x)\ln[\tan\left(\frac{x}{3}\right)]=\color{blue}{-2}$$

17. $$\int_{0}^{\pi/2}\sin^2(x)\ln[\tan(x)]\mathrm dx=\int_{0}^{\pi/2}\sin^4(x)\ln[\tan(x)]\mathrm dx=\color{red}\frac{\pi}{4}$$

18. $$\int_{-\infty}^{+\infty}\frac{\cos(\pi x^2)}{1+2\cosh\left(\frac{x\pi}{\sqrt{3}}\right)}\mathrm dx=\color{green}\sin\left(\frac{\pi}{12}\right)$$

19. $$\int_{0}^{1}\frac{x-3x^3+x^5}{1+x^4+x^8}\ln[-\ln(x)\mathrm dx]=\color{red}\frac{\pi}{\sqrt{3^3}}\cdot\frac{\ln(2)}{2}$$

20. $$4\pi\int_{0}^{\pi/2}\frac{\ln^2[\tan^2(x)]}{(\pi^2+\ln^2[\tan^2(x)])^2}\mathrm dx=\color{green}\ln(2)-\frac{\pi^2}{24}$$

21. $$\int_{0}^{1}\frac{x+x^2}{1-x+x^2}\ln(-\ln x)\mathrm dx=-\gamma-\color{red}{\ln 2 \ln 3}$$

22. $$\int_{0}^{1}x^2\sec(x)[1+\sin(x\pi)-\sin^2(x\pi)]\ln[\sin(x\pi)]=\left(\frac{2}{\pi}\right)^2-\frac{2}{\pi}\cdot\frac{\ln(2)}{\pi}+\frac{\ln(2)}{2}\cdot\frac{\ln(2)}{\pi}$$

23. $$\int_{0}^{\pi/2}\ln^2\left(\frac{\ln^2(\sin x)}{\pi^2+\ln^2(\sin x)}\right)\frac{\mathrm dx}{\tan x}=\color{blue}(2\pi)^2\ln(2)$$

24. $$\int_{0}^{\pi/2}\ln\left(\frac{\ln^2(\sin x)}{\pi^2+\ln^2(\sin x)}\right)\frac{\mathrm dx}{\tan x}=\color{blue}\pi^2$$

25. $$\int_{0}^{\pi/2}\frac{\ln^2(\tan^2 x)}{\pi^2+\ln^2(\tan^2 x)}\mathrm dx=\color{blue}\frac{\pi}{2}(1-\ln 2)$$

26. $$\int_{0}^{1}x^2\cos(x\pi)\ln(\sin x)\mathrm dx=\color{red}\frac{2^2-\ln(2^2)}{\pi^2}$$

27. $$\int_{0}^{1}x\ln(1-x)\ln(1+x)\mathrm dx=\frac{1}{4}-\ln(2)$$

28. $$\int_{0}^{1}x\ln(1-x)\ln^2(1+x)\mathrm dx=\color{brown}\frac{11}{8}-\zeta(2)$$

27. $$\int_{0}^{\infty}\frac{\ln(e^x-1)}{e^x-1}\cdot\frac{\mathrm dx}{x}=\color{blue}\zeta(3)$$

28. $$\int_{0}^{\pi/2}\frac{\cos(2x)}{1+4\sin^2(2x)}\cdot x\ln(\tan x)\mathrm dx=-\left(\frac{\pi}{4}\right)^2 \operatorname{arsinh(2)}$$

29. $$\int_{0}^{\pi/2}\frac{\cos(2x)}{1+\sin^2(2x)}\cdot \ln(\tan x)\mathrm dx=-\frac{\pi}{4}\operatorname{arsinh(2\sqrt{2})}$$

30. $$\int_{0}^{\infty}\frac{\sin^4(\ln x)}{\ln^2 x}\cdot \frac{\mathrm dx}{1+x}=\color{blue}\frac{\pi}{4}$$

31. $$\int_{0}^{\infty}\frac{\sin(\ln^2 x)}{\ln^2 x}\cdot \frac{\mathrm dx}{1+x}=\color{blue}\sqrt{\frac{\pi}{2}}$$

32. $$\int_{0}^{\infty}\frac{\sin(\ln x)}{\ln x}\cdot \frac{\mathrm dx}{1+x}=\color{yellow}\frac{\pi}{2}$$

33. $$\int_{0}^{1}\frac{\ln(1-2x+2x^2)}{x}\mathrm dx=-\frac{\pi^2}{8}$$

34. $$\int_{0}^{1}\frac{\ln(1-2x+2x^2)}{x-1}\mathrm dx=\color{green}\frac{\pi^2}{8}$$

35. $$\int_{0}^{\infty}\sin^2\left(\frac{1}{x}\right)\cdot\frac{\mathrm dx}{(x^2+4)^2}=\color{red}\frac{\pi}{8^2}$$

36. $$\int_{0}^{2\pi}\sin\left(\frac{x}{2}\right)\ln^2\left[\sin\left(\frac{x}{4}\right)\sin\left(\frac{x}{8}\right)\right]\mathrm dx=\frac{27}{2}+\ln(2)+\color{red}\ln^2(2)$$

37. $$\int_{0}^{4\pi}\sin\left(\frac{x}{2}\right)\sin\left(\frac{x}{4}\right)\ln^2\left[\sin\left(\frac{x}{8}\right)\right]\mathrm dx=\frac{\pi}{9}\left(2\ln 12-1\right)$$

38. $$\int_{0}^{\infty}\frac{\sin(x\pi)}{\sinh(x\pi)\tanh(x\pi)}\mathrm dx=\color{blue}\frac{1}{4}$$

39. $$\int_{0}^{\pi/2}\arctan(2\cos^2 x)\cdot\frac{\mathrm dx}{\cos^2 x}=\color{green}\frac{\pi}{\sqrt{\phi}}$$

40. $$\int_{0}^{\pi/2}\arctan(2\tan^2 x)\mathrm dx=\color{blue}\pi\arctan\left(\frac{1}{2}\right)$$

41. $$\int_{0}^{\pi/2}\arctan\left(\sqrt{\frac{\sin(2x)}{\sin^2 x}}\right)\mathrm dx=\color{brown}\pi\arctan\left(\frac{1}{2}\right)$$

42. $$\int_{0}^{\pi/2}\sqrt{\sqrt{x^2+\ln^2(\sin x)}-\ln(\sin x)}\mathrm dx=\color{blue}\frac{\pi}{2}\sqrt{\frac{45}{17}\ln 2}$$

43. $$\int_{0}^{\pi/2}\left(\frac{\sin x}{x}\right)^2\sin^2 x \ln x=\frac{\pi}{4}\color{red}(1-\gamma)$$

44. $$\int_{0}^{\pi/2}\frac{\ln \sin x}{\sin x}\cdot \frac{\ln \cos^2 x}{\cos x}\mathrm dx=-\color{red}\left(\frac{\pi^2}{4!}\right)^2$$

45. $$\int_{0}^{\pi/2}\ln^2\left[\tan^2\left(\frac{x}{2}\right)\right]\mathrm dx=\color{green}\frac{\pi^3}{2}$$

46. $$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{(1+x^2)(\pi+\cos x)}=\color{grey}\frac{\pi}{\sqrt{\pi^2-1}}\left(\frac{\sqrt{\pi-1}+\sqrt{\pi+1}\tanh(1/2)}{\sqrt{\pi+1}+\sqrt{\pi-1}\tanh(1/2)}\right)$$

47. $$\int_{0}^{\infty}\frac{x}{(1+x^2)^2}\cdot\frac{\mathrm dx}{\tanh\left(\frac{x\pi}{2}\right)}=\color{green}\frac{\pi^2}{8}-\color{purple}\frac{1}{2}$$

48. $$\int_{0}^{\infty}\frac{\cos x-\cos(x^2)}{x}\cdot \ln x\mathrm dx=\color{green}\frac{12\gamma^2-\pi^2}{32}$$

49. $$\int_{-\infty}^{+\infty}\left(\frac{x^2-x+\pi}{x^4-x^2+1}\right)^2\mathrm dx=\pi+\color{red}\pi^2+\color{blue}\pi^3$$

50. $$\int_{0}^{\infty}\frac{x\sin^2(x)}{\cosh x+\cos x}\mathrm dx=\color{green}1$$

51. $$\int_{0}^{1}\int_{0}^{1}\frac{2+x^2-y^2}{2-x^2-y^2}\mathrm dx \mathrm dy=\color{green}2G$$

52. $$\int_{0}^{\infty}\frac{1}{x^8+x^4+1}\cdot\frac{1}{x^3+1}\mathrm dx=\color{blue}\frac{5\pi}{12\sqrt{3}}$$

53. $$\int_{0}^{1}\int_{0}^{1}[-\ln(xy)]^s\left(\frac{1}{\ln(xy)}+\frac{1}{1-xy}\right)\mathrm dx \mathrm dy=\color{green}\Gamma(s+2)\left(\zeta(s+2)-\frac{1}{s+1}\right)$$

54. $$4\int_{0}^{\infty}\frac{e^{-x}+x-1}{e^{2x}-e^{-2x}}\cdot\frac{\mathrm dx}{x}=\gamma+\color{red}\ln\frac{16\pi^2}{\Gamma^4(1/4)}$$

55. $$\int_{0}^{1}\left(\frac{x}{1-x}\right)\ln\left(\frac{x}{1-x}\right)\ln(1+x-x^2)\mathrm dx=\color{purple}\frac{2}{5}\zeta(3)$$

56. $$\int_{0}^{1}\left(\frac{x}{1-x}\right)\ln\left(\frac{x}{1-x}\right)\ln\left(\frac{1+x-x^2}{x}\right)\mathrm dx=\color{purple}\frac{1+\sqrt{2}}{\pi}\color{brown}\zeta(3)$$

57. $$\int_{0}^{1}\frac{\arctan(2x)}{\sqrt{x(1-x)}}\mathrm dx=\color{purple}\pi\arccot\sqrt{\phi}$$

58. $$\int_{0}^{1}\frac{\arctan(2\sqrt{2}x)}{\sqrt{x(1-x)}}\mathrm dx=\color{violet}\left(\frac{\pi}{2}\right)^2$$

59. $$\int_{0}^{1}\frac{\arccot(x)}{\sqrt{x(1-x^2)}}\mathrm dx=\color{red}\frac{3}{32}\sqrt{2\pi}\cdot\color{green}\Gamma^2(1/4)$$

60. $$\int_{0}^{1}\frac{\arctan(2x)}{\sqrt{x(1-x)}}\frac{\mathrm dx}{x}=\color{purple}\frac{2\pi}{\sqrt{\phi}}$$

61. $$\int_{0}^{1}\frac{\arctan(2\sqrt{2}x)}{\sqrt{x(1-x)}}\frac{\mathrm dx}{x}=\color{blue}2\pi$$

62. $$\int_{0}^{1}\arccot\sqrt{x(1-x)}\mathrm dx=\color{red}\frac{\pi}{\phi^2}$$

63. $$\int_{0}^{1}x\arccot\sqrt{x(1-x)}\mathrm dx=\color{orange}\frac{\pi}{2\phi^2}$$

64. $$\int_{0}^{\infty}\frac{1}{x}\ln\left(\frac{1}{x}\right)\sin(2x)\sin^2(x)\mathrm dx=\color{green}\frac{\pi\gamma}{8}$$

65. $$\int_{-\alpha}^{+\alpha}\arccos\left(\frac{x}{\alpha}\right)\ln\left(x+\alpha\right)\mathrm dx=\color{blue}\alpha \pi\ln\left(\frac{\alpha}{2}\right)$$

66. $$\int_{0}^{\pi/2}\frac{1+2\cos(2x)\ln\tan x}{1+\tan x}\mathrm dx=-\color{red}\frac{\pi}{4}$$

67. $$\int_{0}^{1}\int_{0}^{1}\frac{x(1-x)\cdot y(1+y)}{(1-xy)\ln(xy)}\mathrm dx\mathrm dy=-\color{green}\frac{1}{2}$$

68. $$\int_{0}^{1}\int_{0}^{1}\frac{(\ln x\ln y)^s}{(1-xy)}\mathrm dx\mathrm dy=\color{green}\Gamma^2(1+s)\color{blue}\zeta(2+2s)$$

69. $$\int_{0}^{\infty}\frac{e^x-x-1}{e^{2x}-1}\cdot\frac{\mathrm dx}{x}=\color{red}\frac{1}{2}\left(\ln \pi -\gamma\right)$$

70. $$\int_{0}^{\infty}\frac{\cosh x-1}{e^{ax}+1}\cdot\frac{\mathrm dx}{x}=\color{green}\frac{1}{2}\ln\left(\frac{2a}{\pi}\cdot \tan\left(\frac{\pi}{2a}\right)\right)$$

71. $$\int_{0}^{\infty}\frac{1}{1+x^2}\cdot \frac{x^{\pi}}{1+x^{\pi}}\cdot\frac{\mathrm dx}{1+x^e}=\int_{0}^{\infty}\frac{1}{1+x^2}\cdot \frac{x^{e}}{1+x^{e}}\cdot\frac{\mathrm dx}{1+x^{\pi}}$$

72. $$\int_{0}^{\infty}\frac{x\ln^2(x)}{1+x^2+x^4}\mathrm dx=\color{red}\left(\frac{\pi}{3}\right)^3\cdot\color{green}\frac{2}{3\sqrt{3}}$$

73. $$\int_{0}^{\infty}\left[\sin^2\left(\frac{1}{x}\right)+x\left[\sin\left(\frac{2}{x}\right)-2\sin\left(\frac{1}{x}\right)\right]\right]\cdot \ln x\mathrm dx=\color{red}\frac{\pi}{4}(1-2\ln 2)$$

74. $$\int_{0}^{\infty}\left[3\sin^2\left(\frac{1}{x}\right)+2x\left[\sin\left(\frac{2}{x}\right)-2\sin\left(\frac{1}{x}\right)\right]\right]\cdot \ln x\mathrm dx=\color{blue}\frac{\pi}{2}(\gamma-\ln 2)$$

75. $$\int_{0}^{\infty}\left[2\sin^2\left(\frac{1}{x}\right)+x\left[\sin\left(\frac{2}{x}\right)-2\sin\left(\frac{1}{x}\right)\right]\right]\cdot \ln x\mathrm dx=\color{green}\frac{\pi}{4}(2\gamma-1)$$

76. $$\int_{0}^{\infty}x\left[\cos\left(\frac{1}{x}\right)-x\sin\left(\frac{1}{x}\right)\right]^2\mathrm dx=\color{orange}\frac{1}{4}$$

77. $$\int_{0}^{\infty}\frac{\left[1-\cos\left(\frac{1}{x}\right)\right]^2}{1+x^2}\mathrm dx=\color{blue}\frac{\pi}{2}\cdot\frac{1}{\zeta(3)}$$

78. $$\int_{0}^{\infty}\frac{\left[\cos\left(\frac{1}{x}\right)-x\sin\left(\frac{1}{x}\right)\right]^2}{1+x^2}\mathrm dx=\color{purple}{\pi}$$

79. $$\sum_{k=0}^{n\ge 1}{n \choose k}(-1)^k\int_{0}^{\infty}e^{-x^{1+k}}\cdot \frac{\mathrm dx}{x}=-\frac{\gamma}{1+n}$$

80. $$\int_{0}^{\infty}\frac{\cos x-\cos(2x)}{x^2}\mathrm dx=\color{green}\frac{\pi}{2}$$

81. $$\int_{0}^{\pi/2}\frac{\ln^2(\sin x )}{\cos^2 x}\mathrm dx=\color{red}\pi \ln 2$$

82. $$\int_{0}^{\infty}\frac{\ln(\cosh x )}{\cosh x}\mathrm dx=\frac{1}{2}\color{blue}\pi \ln 2$$

83. $$\int_{0}^{\pi/2}\frac{\ln(\cos x)\ln(\sin x)}{\cos^2x}\mathrm dx=\color{grey}\pi \left(\ln 2-\frac{1}{2}\right)=0.606789...$$

84. $$\int_{0}^{\pi}\cos x\sqrt{\alpha^2-\sin(2x)}\mathrm dx=\color{red}\frac{\alpha^2-1}{2}\operatorname{arctanh}\left(\frac{1}{\alpha}\right)-\color{blue}\frac{\alpha^2+1}{2}\arctan\left(\frac{1}{\alpha}\right)$$

85. $$\int_{0}^{\pi}\cos x\sqrt{\sin(2x)}\mathrm dx=\color{blue}\frac{\pi}{4}(1-\color{red}i)$$

86. $$\int_{0}^{\pi/2}\cos^2x\sqrt{\sin(2x)}\mathrm dx=\color{blue}\frac{1}{\sqrt{2\pi}}\cdot \Gamma^2\left(\frac{3}{4}\right)$$

87. $$\int_{0}^{1}\frac{\ln(1-x^2)}{1+x^2}\cdot \frac{\mathrm dx}{x^2}=\color{green}C-\color{brown}\left(2+\frac{\pi}{4}\right)\color{blue}\ln 2$$

88. $$\int_{0}^{1}\frac{\ln x\ln(1+x)}{1+x}\mathrm dx=-\color{purple}\frac{\zeta(3)}{8}$$

89. $$\int_{0}^{\pi/4}\frac{\cos(2x)}{\sin^2(2x)}\cdot \ln(\sin x)\ln(\cos x) \mathrm dx=\frac{1}{8}\color{green}[\ln(2^2)-\ln^2(2)]$$

90. $$\int_{0}^{\pi/4}\frac{\ln(\sin x)\ln(\cos x)}{\sin^2(2x)} \mathrm dx=\frac{\pi}{8}\color{grey}[\ln(4)-1]$$

91. $$\int_{0}^{\infty}\arctan(x)\ln(1+x^2)\cdot\frac{\mathrm dx}{x^2}=2\color{green}\zeta(2)$$

92. $$\int_{0}^{\infty}\sqrt{x}\arctan(x)\ln(1+x^2)\cdot\frac{\mathrm dx}{x^2}=2^2\sqrt{2}\color{blue}\pi\ln 2$$

93. $$\int_{0}^{\infty}\frac{\sin(x)e^{-x}\ln(x)}{x}\mathrm dx=-\color{orange}\frac{\pi}{8}(2\gamma+\ln 2)$$

94. $$\int_{0}^{\infty}\frac{2-\cos(3x)[\cos^2(2x)+\cos(4x)]}{xe^x}\mathrm dx=\color{purple}\ln(10)$$

95. $$\int_{0}^{\infty}\ln\left(\frac{1-x}{1+x}\right)\ln \left(\frac{1-x^2}{1+x^2}\right)\frac{\mathrm dx}{x}=\color{purple}\pi \color{gree}C$$

96. $$\int_{0}^{\infty}x\left(1-\frac{\sinh x}{\cosh x+\frac{3}{2}}\right)\mathrm dx=\color{green}\zeta(2)+2\color{brown}\ln^2(\phi)$$

97. $$\int_{0}^{\pi/4}\ln\tan\left(x+\frac{\pi}{4}\right)\frac{\mathrm dx}{\cos^2(x)}=\color{blue}2\ln 2$$

98. $$\int_{0}^{\pi/4}\ln^2\left[\tan\left(x+\frac{\pi}{4}\right)\right]\frac{\mathrm dx}{\cos^2(x)}=\color{green}2\zeta(2)$$

99. $$\int_{0}^{\pi/4}\ln^3\left[\tan\left(x+\frac{\pi}{4}\right)\right]\frac{\mathrm dx}{\cos^2(x)}=3^2\color{grey}\zeta(3)$$

100. $$\int_{0}^{\pi/4}\ln^5\left[\tan\left(x+\frac{\pi}{4}\right)\right]\frac{\mathrm dx}{\cos^2(x)}=15^2\color{brown}\zeta(5)$$