Answer
$$
\int_{0}^{1 / 2} \frac{x e^{2 x}}{(1+2 x)^{2}} d x=\frac{1}{8} e-\frac{1}{4}.
$$
where $C$ is constant.
Work Step by Step
$$
\int_{0}^{1 / 2} \frac{x e^{2 x}}{(1+2 x)^{2}} d x
$$
first, we integrate$
\int \frac{x e^{2 x}}{(1+2 x)^{2}} d x
$ by parts, with
$$
\left[\begin{array}{ll}{u=x e^{2 x},} & {d v=\frac{dx }{(1+2 x)^{2}}} \\ {d u=\left(x \cdot 2 e^{2 x}+e^{2 x} \cdot 1\right) d x, } & {v=-\frac{1}{2} \cdot \frac{1}{1+2 x}}\end{array}\right]
$$
so
$$
\begin{aligned}
\int \frac{x e^{2 x}}{(1+2 x)^{2}} d x &=-\frac{1}{2} \cdot \frac{x e^{2 x}}{1+2 x}-\int\left[-\frac{1}{2} \cdot \frac{e^{2 x}(2 x+1)}{1+2 x}\right] d x \\
&=-\frac{x e^{2 x}}{4 x+2}+\frac{1}{2} \cdot \frac{1}{2} e^{2 x}+C \\
&=e^{2 x}\left(\frac{1}{4}-\frac{x}{4 x+2}\right)+C.
\end{aligned}
$$
Thus,
$$
\begin{aligned}
\int_{0}^{1 / 2} \frac{x e^{2 x}}{(1+2 x)^{2}} d x &=\left[e^{2 x}\left(\frac{1}{4}-\frac{x}{4 x+2}\right)\right]_{0}^{1 / 2} \\
&=e\left(\frac{1}{4}-\frac{1}{8}\right)-1\left(\frac{1}{4}-0\right) \\
&=\frac{1}{8} e-\frac{1}{4}.
\end{aligned}
$$
where $C$ is constant.
