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I stack with following question. $\int \frac {xe^2}{(1+2x)^2}dx$

I think I need to use $uv-\int vdu$ to evaluate this function but I couldn't see which would be $u$ and $v$

If you have any idea could you post it here ?

Thank you !

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Do you really mean $e^2$? That's a constant and doesn't matter. Just make the substitution $1+2x=u$. –  André Nicolas Feb 12 '12 at 20:33
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up vote 6 down vote accepted

I think there is a little simpler than integration by parts. We write $$\frac{e^2x}{(1+2x)^2}=\frac{e^2}2\frac{2x+1-1}{(2x+1)^2}=\frac{e^2}2\left(\frac 1{2x+1}-\frac 1{(2x+1)^2}\right)=\frac{e^2}4\left(\frac 2{2x+1}-\frac 2{(2x+1)^2}\right),$$ so $$\int\frac{e^2x}{(1+2x)^2}dx=\frac{e^2}4\left(\ln(2x+1)+\frac 1{2x+1}\right)+K,$$ where $K$ is a real constant.

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Beat me to it! +1. –  user21436 Feb 12 '12 at 20:33
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