5.5 The Substitution Rule/27: Difference between revisions
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du &=3{z}^2dz \\[2ex] | du &=3{z}^2dz \\[2ex] | ||
\frac{1}{3}du &={z}^2dz \\[2ex] | \frac{1}{3}du &={z}^2dz \\[2ex] | ||
\end{align} | |||
</math> | |||
<math> | |||
\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] | |||
&= -\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{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 16:15, 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)
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} \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{align} }