∫ z 2 1 + z 3 3 d z {\displaystyle \int {\cfrac {z^{2}}{\sqrt[{3}]{1+z^{3}}}}dz}
u = 1 + z 3 d u = 3 z 2 d z 1 3 d u = z 2 d z {\displaystyle {\begin{aligned}u&=1+{z}^{3}\\[2ex]du&=3{z}^{2}dz\\[2ex]{\frac {1}{3}}du&={z}^{2}dz\\[2ex]\end{aligned}}}
∫ z 2 1 + z 3 3 d z = 1 3 ∫ 1 u 3 d u = 1 3 ∫ u − 1 3 d u = − 1 3 ( 3 2 u 2 3 ) = 3 6 u 2 / 3 = 1 2 ( 1 + z 3 ) 2 3 + C {\displaystyle {\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}}}
![{\displaystyle \int {\cfrac {z^{2}}{\sqrt[{3}]{1+z^{3}}}}dz}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a16762723c154c008294ac839f1997ba1455d543)
![{\displaystyle {\begin{aligned}u&=1+{z}^{3}\\[2ex]du&=3{z}^{2}dz\\[2ex]{\frac {1}{3}}du&={z}^{2}dz\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a2fa06923c4c79966cfebb815eb85a2321de6418)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/8cf3ab6276fae0f2c5fb9ea43317348c08d2fb57)