5.5 The Substitution Rule/27

$\int {\cfrac {z^{2}}{\sqrt[{3}]{1+z^{3}}}}dz$

${\begin{aligned}u&=1+{z}^{3}\\[2ex]du&=3{z}^{2}dz\\[2ex]{\frac {1}{3}}du&={z}^{2}dz\\[2ex]\end{aligned}}$

${\begin{aligned}\int {\cfrac {z^{2}}{\sqrt[{3}]{1+z^{3}}}}dz&={\frac {1}{3}}\int {\frac {1}{\sqrt[{3}]{u}}}du={\frac {1}{3}}\int {u}^{-{\frac {1}{3}}}du\\[2ex]&=-{\frac {1}{3}}({\frac {3}{2}}{u}^{\frac {2}{3}})={\frac {3}{6}}{u}^{2/3}\\[2ex]&={\frac {1}{2}}{1+{z}^{3}}^{\frac {2}{3}}+C\end{aligned}}$