<s>[INST] Integrate $\int$ $\frac{x^3dx}{\sqrt{1+x^2}}$ Use $u$ substitution:
$u = 1 + x^2$, $du = 2xdx$ , $dx =\frac {du}{2x}$
Write the given integral in terms of : $u$, $du$ and $dx$ 
$$\frac {1}{2}\int \frac {x^4}{x\sqrt{u}} du \implies \frac{1}{2} \int \frac {u+x^2 - 1}{x \sqrt{u}}$$
This works right I double checked my work. But it looks awfully complicated for a calculus problem than the other ones I have done.
 [/INST] Set $\sqrt{1+x^2}=u\implies 1+x^2=u^2, xdx=u du$
$$\int\frac{x^3}{\sqrt{1+x^2}}dx=\int\frac{u^2-1}uudu$$
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