User:EML4500.f08.JAMAMA/FE/ex45

= HW 2.4 =

Problem Statement
Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

Do the following problems to lead this result

1) Show $$\int{log(x) }dx = xlog|x| - x $$

2) Show $$\int{x log(x)}dx = \frac{1}{2}x^2\left[log(x)-\frac{1}{2}\right]$$

3) Find $$\int{\frac{x^2}{1+cx}}dx$$

4) Find $$\int{\frac{x^2}{a+bx}}dx$$

5) Find the exact solution of $$ \displaystyle u(x) $$ for, $$ \displaystyle            \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$

6) Plot $$ \displaystyle u(x) $$

1) Show $$\int{logx}dx = xlogx - x $$
Using Integration By Parts, (4.1) Let

(4.2)

Entering values from eqns 4.2 into eqn. 4.1 yields

$$xlogx-\int 1dx$$

2) Show $$\int{x logx}dx = \frac{1}{2}x^2\left[logx-\frac{1}{2}\right]$$
Using Integration by parts, let

(4.3) Now entering values from eqns 4.3 into eqn. 4.1 yields

$$\frac{x^2}{2}logx - \int{\left(\frac{x^2}{2}\right) \left(\frac{1}{x}\right)}dx $$

$$\frac{x^2}{2}logx- \frac{x^2}{4}$$

3) Find $$\int \frac{x^2}{1+cx}$$
Using Integration by parts, let (4.4)

Now, entering values from eqns 4.4 into eqn 4.1 yields, (4.5)

Now, integration by parts must be done again. Let (4.6)

The equation now takes the form (4.7)

Entering values from eqns 4.6 into 4.7 yields $$ \displaystyle \frac{x^{2}}{c} log(1+cx) - \left[ {\left( \frac{x^2}{c} log(1+cx)- \frac{1}{c^3}log(1+cx)- \frac{x^2 }{2c} + \frac{x}{c^2} \right)} {\left( 2x \right)} - \int {\left(2dx \right)}\right] $$

Distributing terms,

(4.8)

Solving each integral in 4.8 individually

$$ \displaystyle \begin{align} & {2\int{\frac{x^2}{c}log(1+cx)}\ dx} = \frac{2x^3}{3c}\log(1+cx)+\frac{2}{3c^4}\log(1+cx)-\frac{2x^3}{9c}+\frac{x^2}{3c^2} -\frac{2x}{3c^3}\\ & {-2\int{\frac{1}{c^3}log(1+cx)}\ dx} = -\frac{2x}{c^3}log(1+cx) - \frac{2}{c^4}log(1+cx) + \frac{2x}{c^3} \\ & {\int{\frac{x^2}{c}}\ dx} = \frac{x^3}{3c} \\ & {2\int{\frac{x}{c^2}}\ dx} = \frac{x^2}{c^2} \end{align} $$  (4.9)

Entering values for eqns 4.9 into eqn. 4.8 yields

Which simplifies to

When c=1, the problem further simplifies to

4) Find $$\int \frac{x^2}{a+bx}$$
Using integration by parts, let

(4.10)

Now, entering values from eqns 4.10 into eqn 4.1 yields,

(4.11)

Now, integration by parts must be done again. Let

(4.12)

The equation now takes the form

(4.13)

Entering the values from 4.12 into 4.13 yields,

Distributing terms,

(4.14)

Solving each integral in 4.14 individually,

$$ \displaystyle \begin{align} &\frac{2}{b}\int{xlog(a+bx)}dx = \frac{x^2}{b}\log(a+bx)-\frac{a^2}{b^3}\log(a+bx)-\frac{x^2}{2b}+\frac{ax}{b^2} \\ &\frac{2a}{b^2}\int{log(a+bx)}dx = \frac{2ax}{b^2}log(a+bx)+\frac{2a^2}{b^3}log(a+bx) - \frac{2ax}{b^2} \\ &\frac{2}{b}\int{x}dx = -\frac{x^2}{b} &\end{align} $$  (4.15)

Substituting eqns 4.15 into 4.14 yields

Which simplifies to

 (4.16)

5) Find the exact solution for
$$ \displaystyle \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$  ;    $$  \forall x \in ] 0,1 [ $$

With the following Boundary Conditions:

Essential: $$ u(1)=4 $$

 (4.17) Natural:$$ -\frac{d}{dx} u(x=0) =6 $$

 (4.18)

$$ \frac{d}{du} [(2+3x)\frac{du}{dx}]=-5x $$  (4.19) Integrate both sides of eqn. 4.19,

$$ \int d [(2+3x)\frac{du}{dx}] = \int-5x dx$$

$$ (2+3x)\frac{du}{dx}=\frac{-5x^2}{2}+c_1$$

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{c_1}{2+3x}$$

 (4.20) Apply Natural B.C., eqn. 4.18, to find $$ c_1 $$,

$$-\frac{du}{dx}(x=0) =6 \implies -6= 0+ \frac{c_1}{2} $$

$$c_1=-12$$  (4.21) Enter value from eqn. 4.21 into eqn 4.20

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{-12}{2+3x}$$

$$du=\frac{-5x^2}{2(2+3x)}dx+\frac{-12}{2+3x}dx$$

 (4.22) Integrate both sides of eqn. 4.22

$$\int du=\int\frac{-5x^2}{2(2+3x)}dx+\int\frac{-12}{2+3x}dx$$

$$u=-\frac{5}{2}\int \frac{x^2}{2+3x}dx-12\int \frac{dx}{2+3x}$$

$$u=\frac{-5}{2}[\frac{x^2}{6}-\frac{2x}{9}+\frac{4log(2+3x)}{27}]-12[\frac{log(2+3x)}{3}]+c_2$$

$$u=\frac{-5x^2}{12} +\frac{5x}{9}-\frac{98}{27}log(2+3x)+c_2$$  (4.23) Apply Essential B.C., eqn 4.17, to find

$$ u(1)=4 \implies 4=\frac{-5}{12} +\frac{5}{9}-\frac{98}{27}log(2+3x)+c_2$$

$$c_2=-1.98055$$  (4.24)

Enter the value from eqn 4.24 into eqn. 4.23 to obtain exact solution,

= HW 2.5 =