User:Eml5526.s11.team6.joglekar/hwk2

My Homework
Problem 2.4

=Prove= $$ \int\frac{x^2}{1+x}\, dx=\frac{x^2}{2}-x+log(1+x)+k $$

Solution
Let $$ x^2=u $$

$$ 2x dx=du $$

$$ \frac{1}{1+x}=v $$

$$\frac{(-1)}{(1+x)^2} dx=dv $$

$$ x=w $$

$$dx=dw$$

Rule for Integration by parts:
 * $$\int u v\, dw = u v w - \int u w\, dv - \int v w\, du.\!$$

Therefore $$ \int\frac{x^2}{1+x}\, dx=\ (x^2)*\frac{1}{1+x}*x - \int (x^2)*x*\frac{(-1)}{(1+x)^2}dx- \int\frac{1}{1+x}*x*2x dx $$ $$\int\frac{x^2}{1+x}\, dx=\frac{x^3}{1+x} + \int \frac{x^3}{(1+x)^2}dx - \int\frac{2x^2}{1+x}dx $$

Now Let $$\int\frac{x^2}{1+x}\, dx= I $$

Thus $$\int\frac{x^2}{1+x}\, dx=\frac{x^3}{1+x} + \int \frac{x^3}{(1+x)^2}dx -2I $$

$$3I=\frac{x^3}{1+x} + \int \frac{x^3}{(1+x)^2}dx $$

$$\int \frac{x^3}{(1+x)^2}dx=I_1$$

$$3I=\frac{x^3}{1+x} + I_1$$

$$To find:I_1: $$

let

$$1+x=a $$

$$x=a-1 $$

$$x^3=(a-1)^3 $$

$$\frac{1}{1+x}=\frac{1}{a} $$

Therefore

$$ I_1 =\int\frac{(a-1)^3}{a^2} da $$

$$ I_1=\int\frac{a^3-3a(a-1)-1}{a^2} da $$

$$ I_1=\int a-3\frac{(a-1)}{a}-\frac{1}{a^2}da $$

$$ I_1=\int a da-3\int (1-\frac{1}{a}) da-\int\frac{1}{a^2}du $$

$$ I_1=\frac{a^2}{2}-3a+3log(a)+\frac{1}{a}$$

but a=1+x

Therefore

$$ I_1=\frac{(1+x)^2}{2}-3(1+x)+3log(1+x)+\frac{1}{1+x} $$

Substituting back

$$3I=\frac{x^3}{1+x} + \frac{(1+x)^2}{2}-3(1+x)+3log(1+x)+\frac{1}{1+x} $$

$$3I=\frac{x^3+1}{1+x} + \frac{(1+x)^2}{2}-3(1+x)+3log(1+x) $$

$$I=\frac{x^3+1}{3(1+x)}+\frac{(1+x)^2}{6}- (1+x)+log(1+x) $$

$$I=\frac{(x+1)(x^2-x+1)}{3(x+1)}+\frac{x^2}{6}+\frac{2x}{6}+\frac{1}{6}-1-x+log(1+x) $$

$$I=\frac{x^2}{3}-\frac{x}{3}+\frac{1}{3}+\frac{x^2}{6}+\frac{2x}{6}+\frac{1}{6}-1-x+log(1+x) $$

$$I=\frac{x^2}{2}-x+log(1+x)-\frac{(-1}{2} $$

$$I=\frac{x^2}{2}-x+log(1+x)+k $$

2.4.5
Given:

$$\frac{d}{dx}[(2+3x)*\frac{du}{dx}]+5x = 0 for all x - >0,1< $$

$$ u(1)=4, -\frac{du}{dx} (x=0) = 6 $$

To find:

Expression for u(x):

Solution:

$$ \frac{d}{dx}[(2+3x)\frac{du}{dx}]+5x=0 $$

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

Integrating with respect to x

$$ (2+3x) \frac{du}{dx}= \frac{-5x^2}{2}+k1 $$

$$ \frac{du}{dx}= \frac{-5x^2}{2(2+3x)}+\frac{k1}{2+3x} $$

Integrating again with respect to x

$$ u(x)= \frac{-5}{2}[\frac{2*log(3x+2)+3x(x-4)}{3^3*2}]+k1[\frac{ln(2+3x)}{3}]+k2 $$

To find the constants, we use the given boundary conditions:

$$ \frac{du}{dx}= \frac{-5*x^2}{2(2+3x)}+\frac{k1}{(2+3x)} $$

$$ but \frac{-du}{dx}(x=0)=6 $$

Substituting x=1 in the above equation,

k1= -12

Substituting k1=-12 and simplying, we get:

$$ u(x)= \frac{-118}{27}*log(2+3x)-\frac{5*x^2}{12}+\frac{5}{9}x+k2 $$

Also u(x)=4 at x=1

Substituting x=1 in the expression for u(x);

k2= 10.896

Thus final expression for u(x) is

$$ u(x)=\frac {-118}{27}*log(2+3x)-5*\frac{x^2}{12}+\frac{5}{9}x+10.896 $$