5.5 The Substitution Rule/27: Difference between revisions
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\begin{align} | \begin{align} | ||
\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] | ||
Revision as of 16:26, 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)
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &=1+{z}^3 \\[2ex] du &=3{z}^2dz \\[2ex] \frac{1}{3}du &={z}^2dz \\[2ex] \end{align} }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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}}}