Evaluating $\int_{} \frac{xe^{2x}}{(1+2x)^2}dx$ via integration by parts

Hint:

Let $u=xe^{2x}$, $\text dv=\dfrac{\text dx}{(1+2x)^2}$.

Result: $$\dfrac{\mathrm{e}^{2x}}{8x+4} + C$$

EDIT: More steps.

\begin{eqnarray*} \int \frac{xe^{2x}}{(1+2x)^2} \ \text dx &=& \left|\begin{array}{2} u=xe^{2x} & \text dv=\dfrac{\text d x}{(1+2x)^2} \\ \text du = (1+2x)e^{2x}\ \text dx & v=-\dfrac{1}{2(1+2x)} \end{array}\right| = \\ &=& -\frac{xe^{2x}}{2(1+2x)} + \frac{1}{2} \int e^{2x}\ \text dx = \\ &=&-\frac{xe^{2x}}{2(1+2x)} + \frac{1}{4}e^{2x} + C = \\ &=& e^{2x}\left( -\frac{x}{2(1+2x)}+\frac{1}{4} \right) + C = \boxed{\frac{e^{2x}}{8x+4} + C} \end{eqnarray*}

$u=1+2x$:

$$\require{cancel} \int\frac{xe^{2x}}{(1+2x)^2}\,dx= \frac{1}{4}\int\frac{(1+2x-1)e^{2x+1-1}}{(1+2x)^2}\frac{d}{dx}(1+2x)\,dx=\\ \frac{1}{4}\int\frac{(u-1)e^{u-1}}{u^2}\,du= \frac{1}{4e}\int\left(\frac{ue^{u}}{u^2}-\frac{e^{u}}{u^2}\right)\,du=\\ \frac{1}{4e}\int\left(\frac{e^{u}}{u}-\frac{e^{u}}{u^2}\right)\,du= \frac{1}{4e}\left(\int e^{u}\frac{1}{u}\,du-\int e^{u}\frac{1}{u^2}\,du\right)=\\ \frac{1}{4e}\left(\int e^{u}(\ln{u})'\,du+\int e^{u}\left(\frac{1}{u}\right)'\,du\right)=\\ \frac{1}{4e}\left(e^{u}\ln{u}-\int e^{u}\ln{u}\,du+\frac{e^u}{u}-\int e^{u}\frac{1}{u}\,du\right)=\\ \frac{1}{4e}\left(e^{u}\ln{u}-\int e^{u}\ln{u}\,du+\frac{e^u}{u}-\int e^{u}(\ln{u})'\,du\right)=\\ \frac{1}{4e}\left(\cancel{e^{u}\ln{u}}-\cancel{\int e^{u}\ln{u}\,du}+\frac{e^u}{u}-\cancel{e^{u}\ln{u}}+\cancel{\int e^{u}\ln{u}\,du}\right)=\\ \frac{1}{4e}\frac{e^u}{u}=\frac{1}{4e}\frac{e^{1+2x}}{1+2x}=\frac{e\cdot e^{2x}}{4e(1+2x)}=\frac{e^{2x}}{4+8x}+C. $$

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