<s>[INST] Integrating $\int\frac{xe^{2x}}{(1+2x)^2} dx$ I need help in integrating 
$$\int \frac{x e^{2x}}{(1+2x)^2} dx$$
I used integration by parts to where $u=xe^{2x}$ and $dv=\frac{1}{(1+2x)^2}$ to obtain
$$=\frac{-1}{2(1+2x)}\left[e^{2x}+2xe^{2x}\right] + \frac{1}{2}\int \frac{1}{(1+2x)}\left(2e^{2x}+\frac{d}{dx}\left(2xe^{2x}\right)\right)dx$$
Which is pretty long, which made me question whether I'm doing it correctly or not. Are there any other ways to properly evaluate the integral?
Do note though that I already know the answer to this one since it was given in a textbook, I just don't know how to arrive at such an answer.
$$\int \frac{xe^{2x}}{(1+2x)^2}dx = \frac{e^{2x}}{4+8x}$$
 [/INST] \begin{align}
  & \int \frac{x e^{2x}}{(1+2x)^2} dx \\
= & -\frac{1}{2}\int xe^{2x}d\left(\frac{1}{1 + 2x}\right) \\
= & -\frac{xe^{2x}}{2 + 4x} + \frac{1}{2}\int \frac{e^{2x} + 2xe^{2x}}{1 + 2x}dx \\
= & -\frac{xe^{2x}}{2 + 4x} + \frac{1}{2}\int e^{2x}dx \\
= & -\frac{xe^{2x}}{2 + 4x} + \frac{1}{4}e^{2x} \\
= & \frac{e^{2x}}{4 + 8x} 
\end{align}
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