- #1

QaH

- 6

- 0

## Homework Statement

[tex]\int\frac{x^2}{\sqrt{x^2+4}}dx[/tex]

## Homework Equations

n/a

## The Attempt at a Solution

Letting [tex]x=2tan\theta[/tex] and [tex]dx=2sec^2\theta d\theta[/tex]

[tex]\int\frac{x^2}{\sqrt{x^2+4}}dx=\int\frac{4tan^2\theta}{\sqrt{4+4tan^2\theta}}2sec^2\theta d\theta=\int\frac{8tan^2\theta sec^2\theta}{\sqrt{4(1+tan^2\theta)}}d\theta=4\int\frac{tan^2\theta sec^2\theta}{sec\theta}d\theta=[/tex][tex]4\int tan^2\theta sec\theta d\theta=4\int(sec^2\theta -1)sec\theta d\theta=4\int (sec^3\theta-sec\theta)d\theta=[/tex][tex]4\int sec^3\theta d\theta-4\int sec\theta d\theta=-4\int\frac{cos\theta}{cos^2\theta}d\theta+4\int\frac{cos\theta}{cos^4\theta}d\theta= [/tex][tex]-4\int\frac{cos\theta}{1-sin^2\theta}d\theta+4\int\frac{cos\theta}{(cos^2\theta)^2}d\theta=-4\int\frac{cos\theta}{1-sin^2\theta}d\theta+4\int\frac{cos\theta}{(1-sin^2\theta)^2}d\theta[/tex]

now letting [tex]u=sin\theta[/tex]and[tex]du=cos\theta d\theta[/tex][tex]-4\int\frac{1}{1-u^2}du+4\int\frac{1}{(1-u^2)^2}du=-4\int\frac{1}{(1-u)(1+u)}du+4\int\frac{1}{(1-u^2)^2}du[/tex]

using partial fraction decomposition I get

[tex]\frac{1}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}[/tex] multipying both sides by [tex]1-u^2[/tex][tex]1=A(1+u)+B(1-u)[/tex] Letting u=1 we get that [tex]A=\frac{1}{2}[/tex] now letting u=-1 we get that [tex]B=\frac{1}{2}[/tex] Then [tex]\frac{1}{(1-u)(1+u)}=\frac{1}{2(1-u)}+\frac{1}{2(1+u)}[/tex]

and [tex]-4\int\frac{1}{(1-u)(1+u)}du[/tex] becomes [tex]-4\int\frac{1}{2}(\frac{1}{1-u}+\frac{1}{1+u})du=-2(\ln\mid1+u\mid-\ln\mid1-u\mid+c[/tex]note that [tex]u=sin\theta=\frac{x}{\sqrt{x^2+4}}[/tex][tex]-2(\ln\mid1+\frac{x}{\sqrt{x^2+4}}\mid-\ln\mid1-\frac{x}{\sqrt{x^2+4}}\mid)+c=-2\ln\mid1+\frac{x}{\sqrt{x^2+4}}\mid+2\ln\mid1-\frac{x}{\sqrt{x^2+4}}\mid+c=\ln\mid(1-\frac{x}{\sqrt{x^2+4}})^2\mid-\ln\mid(1+\frac{x}{\sqrt{x^2+4}})^{2}\mid+c=\ln\mid\frac{(1-\frac{x}{\sqrt{x^2+4}})^2}{(1+\frac{x}{\sqrt{x^2+4}})^2}\mid+c[/tex]

and our original integral becomes [tex]\ln\mid\frac{(1-\frac{x}{\sqrt{x^2+4}})^2}{(1+\frac{x}{\sqrt{x^2+4}})^2}\mid+4\int\frac{1}{(1-u^2)^2}du[/tex]

now this is where I am stuck because I can't seem to figue out partial fraction decomposition on [tex]\frac{1}{(1-u^2)^2}=\frac{A}{(1-u^2)}+\frac{B}{(1-u^2)^2}[/tex] multiplying through I get that [tex]1=A(1-u^2)+B[/tex]

Letting u=1 B=1 then[tex]1=A(1-u^2)+1[/tex] then for any u A=0 and I am left with [tex]\frac{1}{(1-u^2)^2}=0+\frac{1}{(1-u^2)^2}[/tex] and I still can't integrate.

Please don't just say I am not using the correct partial fraction method because I have searched and searched but can't find a method on this type of fraction.