<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] Here I give you an alternative approach. If we make $x = \sinh(u)$, we get
\begin{align*}
\int\frac{x^{3}}{\sqrt{x^{2}+1}}\mathrm{d}x & = \int\sinh^{3}(u)\mathrm{d}u = \int(\cosh^{2}(u)-1)\sinh(u)\mathrm{d}u\\
& = \frac{\cosh^{3}(u)}{3} - \cosh(u) + K
\end{align*}
Since $x = \sinh(u) = \pm\sqrt{\cosh^{2}(u)-1}$, we conclude that $\cosh(u) = \sqrt{x^{2} + 1}$. Finally, it results that
\begin{align*}
\int\frac{x^{3}}{\sqrt{x^{2}+1}}\mathrm{d}x = \frac{(x^{2}+1)^{3/2}}{3} - \sqrt{x^{2}+1} + K
\end{align*}
</s>