User:Integrals234

$$\int_{0}^{1}(2x-1)\left(\frac{1}{x(1-x)}+\frac{e^{\pi}}{(1+x)(2-x)}+\frac{e^{2\pi}}{(2+x)(3-x)}+\cdots\right)-\frac{1}{\ln(1-x)}+\frac{x^k}{\ln x}=\ln(k+1)$$

$$\int_{0}^{1}(2x-1)\left(\frac{1}{x(1-x)}+\frac{\phi\pi}{(1+x)(2-x)}+\frac{\phi \pi^2}{(2+x)(3-x)}+\frac{\phi \pi^3}{(2+x)(3-x)}+\cdots\right)-\frac{1}{\ln(1-x)}+\frac{x^{e-1}}{\ln x}=1$$

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$$\int_{0}^{\infty}\left(\frac{2}{e^x-e^{-x}}-\frac{1}{x}\right)^2\mathrm dx=\ln\frac{4}{e}$$

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(z-x^y)(x-2y)\frac{\ln x}{1-x}\mathrm dx \mathrm dy \mathrm dz=2\gamma-1$$

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(z-x^y)[4(y-x)+5(z^3-x^y)]\frac{\ln x}{1-x}\mathrm dx \mathrm dy \mathrm dz=-4\gamma$$

$$\int_{0}^{1}\left(\frac{1}{x}+\frac{x^{n-1}}{\ln(1-x^n)}\right)\mathrm dx=\frac{\gamma}{n}$$

$$\int_{0}^{1}\left(\frac{1}{x}+\frac{1}{x+1}+\cdots\frac{1}{x+m-1}+\frac{x^{n-1}}{\ln(1-x^n)}\right)\mathrm dx=\frac{\gamma}{n}+\ln(m)$$

$$\int_{0}^{1}\left(\frac{1}{1-x}+\frac{x^{n}}{\ln(x)}\right)\mathrm dx =\gamma+\ln(1+n)$$

$$\int_{0}^{1}\left(\frac{1}{1-x}+\frac{1}{2-x}+\cdots+\frac{1}{m-x}+\frac{x^{n-1}}{\ln(x)}\right)\mathrm dx =\gamma+\ln(nm)$$

$$\int_{0}^{1}\left(\frac{1}{1-x^a}+\frac{x^{b}}{\ln(x^a)}\right)\mathrm dx =F(a,b)$$

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

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

$$\int_{0}^{1}\int_{0}^{1}\left(\frac{x}{(1+x)(1+y)(1+xy)}+\frac{1}{\ln(xy)}\right)\mathrm dx \mathrm dy=\frac{\pi^2}{24}-1$$

$$\int_{0}^{1}\int_{0}^{1}\left(\frac{3}{(1+x)(1+y)(1-xy)}+\frac{3y^3}{\ln(xy)}\right)\mathrm dx \mathrm dy=\ln 2$$

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$$\int_{0}^{1}\int_{0}^{1}\left(\frac{x^n}{1-xy}+\frac{x-y}{\ln(xy)}\right)\mathrm dx \mathrm dy=\frac{H_n}{n}$$

$$\int_{0}^{1}\int_{0}^{1}\left(\frac{1}{1-xy}+\frac{x-y}{\ln(xy)}\right)\mathrm dx \mathrm dy=\zeta(2)$$

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

$$\int_{0}^{\infty}\arctan(3\sqrt{x})\left(\frac{1}{x}-\frac{x}{2^2+x^2}\right)\mathrm dx=\pi\ln 5$$

$$\int_{0}^{\infty}\frac{x}{(2n-1)^2+x^2}\cdot \frac{\tanh\left(\frac{\pi}{2}x\right)}{(2n)^2+x^2}\mathrm dx=\frac{2H^{'}_{2n}+\frac{1}{n}-\ln(4)}{4n-1}$$

$$\int_{0}^{\infty}\left(\frac{1}{x}-\frac{x}{(2n-1)^2+x^2}\right)\tanh\left(\frac{\pi}{2}x\right)\mathrm dx=H_{n-1}+\ln(4)$$

$$\int_{0}^{\infty}\left(\frac{1}{x}-\frac{x}{(2n)^2+x^2}\right)\tanh\left(\frac{\pi}{2}x\right)\mathrm dx=2H_{2n}-H_n$$

$$\int_{0}^{1}\int_{0}^{1}\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2}{4}$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{x}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2}{8}$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{x}\right)\ln\left(\frac{1}{y}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2-4}{8}$$

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{1}{x}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2+4}{8}$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{xy}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2}{4}$$

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

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{1}{x}\right)\ln^2\left(\frac{1}{y}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{3\pi^2-16}{8}$$

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{x}{y}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=2$$

$$\int_{0}^{1}\int_{0}^{1}y\ln\left(\frac{y}{x}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\ln 2}{2}$$

$$\int_{0}^{1}\int_{0}^{1}y\ln^2\left(\frac{y}{x}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{3+\ln 2}{4}$$

$$\int_{0}^{1}\int_{0}^{1}\arctan\left(\frac{y}{x}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln\left(\frac{1}{x}\right)-\ln\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=2\beta(3)$$

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$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{y}\right)\frac{\left(\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)\right)^2}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=2\beta(3)$$

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{1}{y}\right)\frac{\left(\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)\right)^2}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\zeta(2)$$

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{1}{xy}\right)\frac{\left(\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)\right)^2}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=4\beta(3)$$

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$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{x}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)-\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2}{8}$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{x}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2}{16}$$

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{1}{x}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2}{16}$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{y}{x}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2}{8}$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{y}{x}\right)\ln\left(\frac{1}{xy}\right)\frac{\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2}{8}$$

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$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{y}\right)\frac{\left(\ln^2\ln\left(\frac{1}{x}\right)-\ln^2\ln\left(\frac{1}{y}\right)\right)^2}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=8\beta(3)$$

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{1}{y}\right)\frac{\left(\ln^2\ln\left(\frac{1}{x}\right)-\ln^2\ln\left(\frac{1}{y}\right)\right)^2}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=4\zeta(2)$$

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{1}{xy}\right)\frac{\left(\ln^2\ln\left(\frac{1}{x}\right)-\ln^2\ln\left(\frac{1}{y}\right)\right)^2}{\ln^2\left(\frac{1}{x}\right)+\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=16\beta(3)$$

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$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{x}\right)\frac{\left(\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)\right)^2}{\ln^2\left(\frac{1}{x}\right)-\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\zeta(2)$$

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{1}{x}\right)\frac{\left(\ln\ln\left(\frac{1}{x}\right)-\ln\ln\left(\frac{1}{y}\right)\right)^2}{\ln^2\left(\frac{1}{x}\right)-\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\zeta(2)$$

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

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{y}{x}\right)\frac{\ln\ln^2\left(\frac{1}{x}\right)-\ln\ln^2\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)-\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=2$$

$$\int_{0}^{1}\int_{0}^{1}\ln^2\left(\frac{1}{x^3y}\right)\frac{\ln\ln^2\left(\frac{1}{x}\right)-\ln\ln^2\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)-\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=2(\pi^2+1)$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{x^3y}\right)\frac{\ln\ln^2\left(\frac{1}{x}\right)-\ln\ln^2\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)-\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\pi^2$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{x^2y}\right)\frac{\ln\ln^2\left(\frac{1}{x}\right)-\ln\ln^2\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)-\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{3\pi^2}{4}$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{x}\right)\ln\left(\frac{1}{y}\right)\frac{\ln\ln^2\left(\frac{1}{x}\right)-\ln\ln^2\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)-\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2-4}{8}$$

$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1}{x}\right)\ln\left(\frac{1}{xy}\right)\frac{\ln\ln^2\left(\frac{1}{x}\right)-\ln\ln^2\left(\frac{1}{y}\right)}{\ln^2\left(\frac{1}{x}\right)-\ln^2\left(\frac{1}{y}\right)}\mathrm dx \mathrm dy=\frac{\pi^2}{4}$$

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

$$\int_{0}^{1}\int_{0}^{1}2xy\ln(x)\ln(y)\ln(xy)\ln\left(x^{(2n+1)^2}y\right)\ln\left(\frac{x}{y}\right)\frac{\ln\ln^2(x)-\ln\ln^2(y)}{\ln^2(x)-\ln^2(y)}\mathrm dx \mathrm dy=\sum_{k=1}^{n}k$$

$$n\ge1$$

$$\lim_{n \to 0}\frac{1}{n}\left(1+\frac{\zeta(1-n)}{\zeta(1+n)}\right)=2\gamma$$

$$\lim_{n \to 0}\frac{1}{n}\left(1+\frac{\zeta(1+n)}{\zeta(1-n)}\right)=-2\gamma$$

$$\lim_{n \to 0}\frac{1}{n}\left(1-\left(\frac{\zeta(1-n)}{\zeta(1+n)}\right)^{2k}\right)=4k\gamma$$ -

$$\int_{0}^{1}\int_{0}^{1}\frac{(x-1)(2x-y+1)}{(1-x^2y^2)\ln(xy))}\mathrm dx \mathrm dy=\gamma$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(x-1)(2y^2-y+1)}{(1-x^2y^2)\ln(xy))}\mathrm dx \mathrm dy=\gamma$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(x-1)(2y^2-y-1)}{(1-x^2y^2)\ln(xy))}\mathrm dx \mathrm dy=\ln\frac{\pi}{4}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(x-1)(2y^2-x)}{(1-x^2y^2)\ln(xy))}\mathrm dx \mathrm dy=\frac{1}{2}\ln\frac{\pi}{2}$$

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

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

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

$$\int_{0}^{1}\int_{0}^{1}\frac{(xy-1)(2y^2+xy-x-1)}{(1-x^2y^2)\ln(xy))}\mathrm dx \mathrm dy=\ln\frac{e}{\pi}$$

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

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

$$\int_{0}^{1}\int_{0}^{1}\frac{(xy-1)(2y^2-x^2y+x-y+a)}{(1-x^2y^2)\ln(xy))}\mathrm dx \mathrm dy=\ln\frac{2^a\pi}{2}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(xy-1)(2y^2-x^2y)}{(1-x^2y^2)\ln(xy))}\mathrm dx \mathrm dy=\ln\frac{\pi}{2}$$

-- $$\int_{0}^{1}\int_{0}^{1}\frac{(ay+1)(x-1)}{(1-xy\ln(xy))}\mathrm dx \mathrm dy=1+a-a\gamma$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(ay-1)(x-1)}{(1-xy\ln(xy))}\mathrm dx \mathrm dy=a-(a+1)\gamma$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(y-1)(ax^2+1)}{(1-xy\ln(xy))}\mathrm dx \mathrm dy=\frac{a}{2}\ln 2+\gamma$$

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

$$\int_{0}^{1}\int_{0}^{1}\frac{(y^2+y-2)(x^2+1)}{(1-xy\ln(xy))}\mathrm dx \mathrm dy=\gamma+\frac{3+\ln 2}{2}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(y^2+y+a)(x^2-x)}{(1-xy\ln(xy))}\mathrm dx \mathrm dy=\frac{1+a\ln 2}{2}$$

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$$\int_{0}^{1}\int_{0}^{1}\frac{x^4}{\left(\frac{1}{2}-x+x^2\right)\left(\frac{1}{4}-x+x^2\right)\sqrt{4+y^2}}\mathrm dx \mathrm dy=\pi\ln(\phi)$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x^2}{\left(\frac{1}{2}-x+x^2\right)\left(\frac{1}{4}-x+x^2\right)\sqrt{4+y^2}}\mathrm dx \mathrm dy=-4\ln(\phi)$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x^3}{\left(\frac{1}{2}-x+x^2\right)^2\left(\frac{1}{4}-x+x^2\right)\sqrt{4+y^2}}\mathrm dx \mathrm dy=-4\ln(\phi)$$

$$\int_{0}^{1}\int_{0}^{1}\frac{\mathrm dx \mathrm dy}{\left(\frac{1}{2}-x+x^2\right)\sqrt{4-y^2}}=\frac{\pi^2}{6}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x^2}{\left(\frac{1}{2}-x+x^2\right)\sqrt{4-y^2}}\mathrm dx \mathrm dy=\frac{\pi}{6}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x^3}{(2x^2-2x+1)^2\left(x-\frac{1}{2}\right)^2\sqrt{(\phi+y^2)^3}}\mathrm dx \mathrm dy=-\frac{1}{\phi^2}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x^3}{(2x^2-2x+1)^2\left(x-\frac{1}{2}\right)^2\sqrt{(a+by^2)^3}}\mathrm dx \mathrm dy=-\frac{1}{a\sqrt{a+b}}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x^3}{(2x^2-2x+1)^2\left(x-\frac{1}{2}\right)^2\sqrt{(a-by^2)^3}}\mathrm dx \mathrm dy=-\frac{1}{a\sqrt{a-b}}$$