Integrating $x^3\sqrt{ x^2+4 }$ Trying to integrate $\int x^3 \sqrt{x^2+4 }dx$, I did the following 
$u = \sqrt{x^2+4 }$   , $du = \dfrac{x}{\sqrt{x^2+4}} dx$
$dv=x^3$    ,  $v=\frac{1}{4} x^4$
$\int udv=uv- \int vdu$
$= \frac{1}{4} x^4\sqrt{x^2+4 } - \int \frac{1}{4} x^4\dfrac{x}{\sqrt{x^2+4}} dx$   ---> i'm stuck here
$\int \dfrac{1}{4 x^4} \dfrac{x}{\sqrt{x^2+4}} dx$   ---> i'm stuck here please help

Use substitution method
First, let $u = x^2$
Then you will have $du = 2x dx$
$$ \int\,x^3\sqrt{x^2+4}\,dx = \dfrac{1}{2}\int\,u\sqrt{u+4}\,du $$
Then let $s = u+4$, which implies $ds=du$
You should be able to get the answer
If you are right, you should get the following:$\frac{ 1}{5} (x^2+4)^\frac {5}{2} -\frac{ 4}{3} (x^2+4)^\frac {3}{2} +C $, where $C$ is a constant.
