Answer
$$\int \frac{x e^{2 x}}{(2 x+1)^{2}} d x =-\frac{x^{2} e^{x^{2}}}{2\left(x^{2}+1\right)}+\frac{1}{2} e^{x^{2}} +c$$
Work Step by Step
Since $$
\int \frac{x^{3} e^{x^{2}}}{\left(x^{2}+1\right)^{2}} d x
$$
integrate by parts , let
\begin{align*}
u&= x^2e^{ x^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\frac{xdx}{\left(x^{2}+1\right)^{2}} \\
u&= 2 x e^{x^{2}}\left(x^{2}+1\right) d x\ \ \ \ \ v=-\frac{1}{2\left(x^{2}+1\right)}
\end{align*}
Then
\begin{align*}
\int \frac{x e^{2 x}}{(2 x+1)^{2}} d x&=uv-\int vdu\\
&=-\frac{x^{2} e^{x^{2}}}{2\left(x^{2}+1\right)}+\frac{1}{2} \int e^{x^{2}}(2 x) d x\\
&=-\frac{x^{2} e^{x^{2}}}{2\left(x^{2}+1\right)}+\frac{1}{2} e^{x^{2}} +c
\end{align*}
