<s>[INST] How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^2+1})}^{3/2}-{\frac{1}{2}}{\sqrt {x^2+1}}+C$$for some constant $C$
 [/INST] Just to give another approach, let $x^2+1=u^2$, in which case $xdx=udu$ and we have
$$\int{x^3\over\sqrt{x^2+1}}dx=\int{x^2\cdot xdx\over\sqrt{x^2+1}}=\int{(u^2-1)udu\over u}=\int(u^2-1)du={1\over3}u^3-u+C\\={1\over3}(x^2+1)^{3/2}-(x^2+1)^{1/2}+C$$
The main virtue (if any) of this approach is that it gets rid of the square root symbol for the integration step.
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