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 $C$

You made a small mistake on the last line; integrating $\frac{1}{2}u^{1/2}-\frac{1}{2}u^{-1/2}$ should give $\frac{1}{3}u^{3/2}-u^{1/2}+C$, not $\frac{2}{3}u^{3/2}-\frac{1}{2}u^{1/2}+C$. When you want to double-check an antiderivative, it's worth differentiating it to see if you get the expected integrand.
