<s>[INST] 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
 [/INST] 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.
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