Answer
$$
\int\frac{x^{3}}{\sqrt {x^{2}+1}}d x=\frac{1}{3}(x^{2}+1)^{\frac{1}{2}}[x^{2}-2] +C
$$
where $ C $ is an arbitrary constant.
Work Step by Step
$$
\int\frac{x^{3}}{\sqrt {x^{2}+1}}d x
$$
Let $ u= x^{2}+1$. Then $ x^{2} =u-1$ and $ x dx =\frac{1}{2} du $, so
$$
\begin{split}
\int\frac{x^{3}}{\sqrt {x^{2}+1}}d x & =\int\frac{(u-1)}{\sqrt {u}}(\frac{1}{2} du )\\
& =\frac{1}{2} \int(u^{\frac{1}{2}}-u^{-\frac{1}{2}}) du
\\
& =\frac{1}{2} (\frac{2}{3}u^{\frac{3}{2}}-2u^{\frac{1}{2}}) +C
\\
& = \frac{1}{2} (\frac{2}{3}u^{\frac{3}{2}}-2u^{\frac{1}{2}}) +C\\
&=\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}-(x^{2}+1)^{\frac{1}{2}} +C\\
&=\frac{1}{3}(x^{2}+1)^{\frac{1}{2}}[(x^{2}+1)-3] +C\\
&=\frac{1}{3}(x^{2}+1)^{\frac{1}{2}}[(x^{2}+1)-3] +C\\
&=\frac{1}{3}(x^{2}+1)^{\frac{1}{2}}[x^{2}-2] +C
\end{split}
$$
where $ C $ is an arbitrary constant.