Integrating $\int\frac{xe^{2x}}{(1+2x)^2} dx$

Question

I need help in integrating

$$\int \frac{x e^{2x}}{(1+2x)^2} dx$$

I used integration by parts to where $u=xe^{2x}$ and $dv=\frac{1}{(1+2x)^2}$ to obtain

$$=\frac{-1}{2(1+2x)}\left[e^{2x}+2xe^{2x}\right] + \frac{1}{2}\int \frac{1}{(1+2x)}\left(2e^{2x}+\frac{d}{dx}\left(2xe^{2x}\right)\right)dx$$

Which is pretty long, which made me question whether I'm doing it correctly or not. Are there any other ways to properly evaluate the integral?

Do note though that I already know the answer to this one since it was given in a textbook, I just don't know how to arrive at such an answer.

$$\int \frac{xe^{2x}}{(1+2x)^2}dx = \frac{e^{2x}}{4+8x}$$

Answer 1

\begin{align} & \int \frac{x e^{2x}}{(1+2x)^2} dx \\ = & -\frac{1}{2}\int xe^{2x}d\left(\frac{1}{1 + 2x}\right) \\ = & -\frac{xe^{2x}}{2 + 4x} + \frac{1}{2}\int \frac{e^{2x} + 2xe^{2x}}{1 + 2x}dx \\ = & -\frac{xe^{2x}}{2 + 4x} + \frac{1}{2}\int e^{2x}dx \\ = & -\frac{xe^{2x}}{2 + 4x} + \frac{1}{4}e^{2x} \\ = & \frac{e^{2x}}{4 + 8x} \end{align}

Answer 2

HINT:

$$2\frac{x e^{2x}}{(1+2x)^2}=\dfrac{d[e^{2x}]}{dx}\cdot\dfrac1{1+2x}+e^{2x}\cdot\dfrac{d\left[\dfrac1{1+2x}\right]}{dx}$$

Another way could be:

Set $2x=y$ and write the numerator as $1+y-1$

and use $\dfrac{d\left[\dfrac1y\right]}{dy}=-\dfrac1{y^2}$ to match with

$$\dfrac{d\{e^z\cdot f(z)\}}{dz}=e^z[f(z)+f'(z)]$$

Answer 3

Notice, $$\int\frac{xe^{2x}}{(1+2x)^2}dx$$ Let $1+2x=t\implies 2dx=dt$

$$=\frac{1}{4}\int\frac{(t-1)e^{t-1}}{t^2}dt$$ $$=\frac{1}{4}\int e^{t-1}\left(\frac{1}{t}-\frac{1}{t^2}\right)dt$$

$$=\frac{1}{4}\left[\int \frac{e^{t-1}}{t}-\int \frac{e^{t-1}}{t^2}\right]$$

$$=\frac{1}{4}\left[\frac{e^{t-1}}{t} +\int \frac{e^{t-1}}{t^2}-\int \frac{e^{t-1}}{t^2}\right]$$ $$=\frac{1}{4}\frac{e^{t-1}}{t}$$$$=\frac{e^{2x}}{4(1+2x)}+c$$