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

Hint: Let be $u = x^2$, then $du = 2x\,dx$, and so you got
$$ \int\,x^3\sqrt{x^2+4}\,dx = \dfrac{1}{2}\int\,u\sqrt{u+4}\,du $$
and the let be $t = u+4$, so $u = t-4$, $dt = du$, and
$$ \dfrac{1}{2}\int\,u\sqrt{u+4}\,du = \dfrac{1}{2}\int\,(t-4)\sqrt{t}\,dt $$
you can calculate this integral by yourself.
