Answer
$$\int \frac{x e^{2 x}}{(2 x+1)^{2}} d x =\frac{-xe^{2 x}}{2\left(2x+1\right)}+\frac{1}{4} e^{2 x}+c$$
Work Step by Step
Since $$
\int \frac{x e^{2 x}}{(2 x+1)^{2}} d x
$$
integrate by parts, let
\begin{align*}
u&= xe^{2 x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\frac{ dx }{(2 x+1)^{2}} \\
u&= (2x+1)e^{2 x}dx\ \ \ \ \ \ v=\frac{-1}{2\left(2x+1\right)}
\end{align*}
Then
\begin{align*}
\int \frac{x e^{2 x}}{(2 x+1)^{2}} d x&=uv-\int vdu\\
&=\frac{-xe^{2 x}}{2\left(2x+1\right)}+\int \frac{(2x+1)e^{2 x}}{2\left(2x+1\right)}dx\\
&=\frac{-xe^{2 x}}{2\left(2x+1\right)}+\frac{1}{2}\int e^{2 x} dx\\
&=\frac{-xe^{2 x}}{2\left(2x+1\right)}+\frac{1}{4} e^{2 x}+c
\end{align*}
