Q:

How to evaluate $\int \frac {xe^2}{(1+2x)^2}dx$

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 !

A:

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.

