Answer
$$-\frac{x e^{2 x}}{2(1+2 x)}+\frac{1}{4} e^{2 x}+C$$
Work Step by Step
Given $$\int \frac{x e^{2 x}}{(1+2 x)^{2}} d x$$
Let
\begin{align*}
u&=xe^{2x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\frac{dx}{(1+2x)^2}\\
u&=(2x+1)e^{2x}dx\ \ \ \ \ \ \ \ \ \ \ dv= \frac{-1}{2(1+2x)}
\end{align*}
Then
\begin{align*}
\int \frac{x e^{2 x}}{(1+2 x)^{2}} d x&=-\frac{x e^{2 x}}{2(1+2 x)}+\frac{1}{2} \int \frac{e^{2 x}(2 x+1)}{1+2 x} d x\\
&=-\frac{x e^{2 x}}{2(1+2 x)}+\frac{1}{2} \int e^{2 x} d x\\
&=-\frac{x e^{2 x}}{2(1+2 x)}+\frac{1}{4} e^{2 x}+C
\end{align*}
