Very tough integral problem.

**navynukejones** · April 23rd 2008, 04:38 PM

integrate: x^2dx/x^2+2x+1

I have no idea how to do this one, please help!!

**TheEmptySet** · April 23rd 2008, 04:45 PM

Originally Posted by navynukejones

integrate: x^2dx/x^2+2x+1

I have no idea how to do this one, please help!!

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

let u=x+1 and du=dx x=u-1

so we get

$\int \frac{x^2}{(x+1)^2}dx=\int \frac{(u-1)^2}{u^2}du=$

$\int \frac{u^2-2u+1}{u^2}du=\int \left( 1-\frac{2}{u}+\frac{1}{u^2}\right)du$

$=u-2\ln|u|-\frac{1}{u}+c=(x+1)-2\ln|x+1|-\frac{1}{x+1}+c$

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