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.
$$\int \frac{xe^{2x}}{(1+2x)^2}dx = \frac{e^{2x}}{4+8x}$$
 A: \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}
A: Notice, $$\int\frac{xe^{2x}}{(1+2x)^2}dx$$
Let $1+2x=t\implies 2dx=dt$
$$=\frac{1}{4}\int\frac{(t-1)e^{t-1}}{t^2}dt$$
$$=\frac{1}{4}\int e^{t-1}\left(\frac{1}{t}-\frac{1}{t^2}\right)dt$$
$$=\frac{1}{4}\left[\int \frac{e^{t-1}}{t}-\int \frac{e^{t-1}}{t^2}\right]$$
$$=\frac{1}{4}\left[\frac{e^{t-1}}{t} +\int \frac{e^{t-1}}{t^2}-\int \frac{e^{t-1}}{t^2}\right]$$
$$=\frac{1}{4}\frac{e^{t-1}}{t}$$$$=\frac{e^{2x}}{4(1+2x)}+c$$
A: HINT:
$$2\frac{x e^{2x}}{(1+2x)^2}=\dfrac{d[e^{2x}]}{dx}\cdot\dfrac1{1+2x}+e^{2x}\cdot\dfrac{d\left[\dfrac1{1+2x}\right]}{dx}$$
Another way could be:
Set  $2x=y$ and write the numerator as $1+y-1$
and use $\dfrac{d\left[\dfrac1y\right]}{dy}=-\dfrac1{y^2}$ to match with
$$\dfrac{d\{e^z\cdot f(z)\}}{dz}=e^z[f(z)+f'(z)]$$
