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.
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{2}\int e^{2x}dx$
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{2} \frac{e^{2x}}{2}+C$
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{4} e^{2x}+C$
 A: When you have a short time to compute antiderivatives looking like
$$I=\int \frac {e^{k x}\,P_n(x)} {[Q_m(x)]^r} \,dx$$ think that a possible solution could be of the form
$$\frac {e^{k x}\,R_p(x)} {[Q_m(x)]^{r-1}}$$ where $P,Q,R$ are polynomials of different degrees.
Differentiate both sides to get
$$\frac {e^{k x}} {[Q_m(x)]^r}\,P_n(x)=\frac {e^{k x}} {[Q_m(x)]^r}\left(Q_m(x) \left(k R_p(x)+R_p'(x)\right)+(1-r) R_p(x) Q_m'(x) \right)$$ Comparing the degrees, we then have
$$n=m+p+p(m-1)\implies p=\frac n m-1$$ So, in $n$ is a muliple of $m$, big hope !
Tring for your case $n=1$, $m-1$, then $p=0$ that is to say that $R_p(x)$ is just a constant  $R_p(x)=a$. Ralacing all numbers, we then have
$$11x=(2x+1) (2 a+0)+(1-2)a 2=4a x\implies a=\frac {11}4$$
$$\int \frac{11xe^{2x}}{(1+2x)^2}dx=\frac {11 e^{2x}}{4(1+2x)}$$
As you can notice, I did not use any integration step to get the final result.
A: 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.
