Solve the integral $\int \frac{e^{2x}(2x+1)}{(x+1)^2}$ can you help me with this excercises: 
$$\int \frac{e^{2x}(2x+1)}{(x+1)^2}$$
I  can not get to the answer book:
the answer is $$\frac{e^{2x}}{x+1}$$
How?
 A: Substitute $u=x+1$, you obtain $$\class{steps-node}{\cssId{steps-node-1}{\mathrm{e}^{-2}}}{\displaystyle\int}\dfrac{\left(2u-1\right)\mathrm{e}^{2u}}{u^2}\,\mathrm{d}u$$
Then use linearity of the integral and integrate by part the term $\frac{e^{2u}}{u^2}$. 
$$\int \frac{e^{2u}}{u^2}=\frac{e^{2u}}{u}-\int\frac{2ue^{2u}}{u^2}du$$
Finally :
$$\class{steps-node}{\cssId{steps-node-1}{\mathrm{e}^{-2}}}{\displaystyle\int}\dfrac{\left(2u-1\right)\mathrm{e}^{2u}}{u^2}\,\mathrm{d}u=\frac{e^{2u-2}}{u}$$
$u=x+1$ so $$\frac{e^{2u-2}}{u}=\frac{e^{2x}}{x+1}$$
A: HINT:
$$\int\dfrac{e^{2x}\{2(x+1)-1\}}{(x+1)^2}dx=\int\left(\dfrac1{1+x}\cdot\dfrac{d(e^{2x})}{dx}+e^{2x}\cdot\dfrac{d\left(\dfrac1{1+x}\right)}{dx}\right)dx$$
$$\dfrac{d(e^{ax}f(x))}{dx}=?$$
A: As a thumb rule, if $f(x)$ is a rational function and $f(x)\, e^{kx}$ has an elementary primitive, its primitive is $C+g(x)\,e^{kx}$ with $g(x)$ being a rational function. The proof of this fact was the origin of differential Galois theory with Liouville theorem. In our case, $f(x)=\frac{2x+1}{(x+1)^2}$ has a double pole at $x=-1$, hence $g(x)$ must be something like $\frac{a+bx}{1+x}$.
Now we may find $a,b$ through differentiation:
$$ \frac{d}{dx}\frac{a+bx}{1+x}e^{2x}=\frac{e^{2x}}{(1+x)^2}\left(2bx^2+2(a+b)x+(a+b)\right)$$
leads to $b=0$ and $a=1$:
$$ \int \frac{e^{2x}(2x+1)}{(x+1)^2}\,dx = C+\color{red}{\frac{e^{2x}}{1+x}}.$$
A: Letting $u=e^{2x}(2x+1)$ and  $dv=\displaystyle\frac{1}{(x+1)^2}dx,\;\;$ so $du=e^{2x}(4x+4)dx$ and $v=-\displaystyle\frac{1}{x+1}dx$, gives
$\displaystyle\int\frac{e^{2x}(2x+1)}{(x+1)^2}dx=-\frac{e^{2x}(2x+1)}{x+1}+\int 4e^{2x}dx=-\frac{e^{2x}(2x+1)}{x+1}+2e^{2x}+C=\frac{e^{2x}}{x+1}+C$
