$$\int \frac {xe^{2x}}{(1+2x)^2}dx$$
I've tried setting $u=e^2$ and $dv=\frac {x}{(1+2x)^2}$ but I'm getting a really long partial answer like:
$$\int \frac {xe^2}{(1+2x)^2}dx = e^{2x}\left[\frac {\ln (1+2x)}{4}+ \frac {1}{4(1+2x)^2}\right] - \ldots$$
I integrated $\frac {x}{(1+2x)^2}$ by decomposing the fraction into $\frac {1}{2(1+2x)}- \frac {1}{2(1+2x)^2}$ which would then give the integral as
$$\int \frac {x}{(1+2x)^2}dx = \int \left[ \frac {1}{2(1+2x)} - \frac {1}{2(1+2x)^2}\right]dx$$
But yes I'm still stuck after doing all of this things. Am I on the right path or should I change the values of $u$ and $dv$?
