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
 A: Use Substitution instead. Let $u^2=x^2+4$. Then $u\,du=x\,dx$ and  we end up integrating $(u^2-4)(u^2)$. 
A: 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.
A: 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.
A: As a general rule, whenever dealing with integrands containing $\sqrt{x^2+a^2}$ in their expression, the 
natural substitutions are $x=a\tan t$ or $x=a\sinh u$, since, on one hand, $\sqrt{\tan^2t+1}=\dfrac1{\cos t},$ 
and $\tan't=\dfrac1{\cos^2t},$ transforming our integral into $\displaystyle16\int\frac{\sin^3t}{\cos^6t}dt=-16\int\frac{1-\cos^2t}{\cos^6t}d(\cos t)$ 
whose evaluation is trivial; and, on the other hand, $\sqrt{\sinh^2u+1}=\cosh u$, and $\sinh'u=\cosh u$
yielding $\displaystyle16\int\sinh^3u~\cosh^2u~du=16\int(\cosh^2u-1)\cosh u~d(\cosh u)$, which is also trivial to evaluate.
