<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] Use Substitution instead. Let $u^2=x^2+4$. Then $u\,du=x\,dx$ and  we end up integrating $(u^2-4)(u^2)$. 
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