Exercise:
Compute, $\int{\frac{{{x}^{3}}}{\sqrt[3]{{{x}^{2}}+1}}dx}$
Solution:
Let ${{u}^{3}}={{x}^{2}}+1\Rightarrow 3{{u}^{2}}du=2xdx\Rightarrow
\frac{3}{2}{{u}^{2}}du=dx$ & ${{x}^{2}}={{u}^{3}}-1$
So $\int{\frac{{{x}^{3}}}{\sqrt[3]{{{x}^{2}}+1}}dx=\int{\frac{x{{x}^{2}}}{\sqrt[3]{{{x}^{2}}+1}}dx}=\frac{3}{2}\int{\frac{{{u}^{2}}\left(
{{u}^{3}}-1 \right)du}{u}}}$
$=\frac{3}{2}\int{u\left(
{{u}^{3}}-1 \right)du}$$=\frac{3}{2}\int{{{u}^{4}}du-\frac{3}{2}\int{u\,du}=\frac{3}{2}\times
\frac{{{u}^{5}}}{5}-\frac{3}{2}\times \frac{{{u}^{2}}}{2}}+c$
$=\frac{3}{10}{{\left( \sqrt[3]{{{x}^{2}}+1}
\right)}^{5}}-\frac{3}{4}{{\left( \sqrt[3]{{{x}^{2}}+1} \right)}^{2}}+c$