Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$ Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$
Can somebody check my solution? Thanks!!
Let $u=11xe^{2x}$
then $\frac{du}{dx} = \frac{d}{dx}(11xe^{2x})$
$=11(e^{2x}+2xe^{2x})$
and so $du=11e^{2x}(2x+1)dx$
Now, let $dv=\frac{1}{(2x+1)^2}dx$
Then $\int dv = \int \frac{1}{(2x+1)^2}dx$
To evaluate this, let $r=2x+1$ and so $dr = 2dx$
Then we have $\int dv = \int \frac{1}{r^2}\frac{dr}{2} = \frac{-1}{2r}=\frac{-1}{2(2x+1)}$
So $v=\frac{-1}{2(2x+1)}$
Wow... So that was already a lot of work to find $du$ and $v$ once we chose $u$ and $dv$... Integration by parts totally sucks!!
Anyway, we are now ready to use our VOODOO formula!!
$\int \frac{11xe^{2x}}{(1+2x)^2}dx = \int udv = uv - \int vdu$
$=(11xe^{2x})(\frac{-1}{2(2x+1)}) - \int \frac{-1}{2(2x+1)}(11e^{2x}(2x+1))dx$
Now, there is a $2x+1$ in the denominator and the numerator so the cancel. We also pull the coefficient $\frac{-11}{2}$ outside of the integral, which we can do.
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{2}\int e^{2x}dx$
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{2} \frac{e^{2x}}{2}+C$
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{4} e^{2x}+C$

Your answer is correct, but it can be further simplified:
$$\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{4}e^{2x}$$
$$=\frac{11}{4}e^{2x} \left(\frac{-2x}{2x+1} + 1 \right) =\frac{11}{4}e^{2x} \left(\frac{-2x-1}{2x+1} + \frac{1}{2x+1} + 1 \right)$$
$$=\frac{11}{4}e^{2x} \cdot \frac{1}{2x+1} = \frac{11e^{2x}}{4(2x+1)}$$
which is the same as the answer in Venkat Amith's comment.
