5.5 The Substitution Rule/27: Difference between revisions
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\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] | \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}{3}(\frac{3}{2}{u}^\frac{2}{3}) = \frac{3}{6}{u}^{2/3} \\[2ex] | ||
&= -\cos{(u)} + C \\[2ex] | &= -\cos{(u)} + C \\[2ex] | ||
&= \frac{1}{2}{1+z^3}^\frac{2}{3} + C | &= \frac{1}{2}{1+z^3}^\frac{2}{3} + C | ||
Revision as of 16:27, 7 September 2022
![{\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]&=-\cos {(u)}+C\\[2ex]&={\frac {1}{2}}{1+z^{3}}^{\frac {2}{3}}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/9a4c1fe5d22970bdb08e9a1e1f45d6491e87e2a8)