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 | \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{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 16:38, 7 September 2022
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 \int \cfrac{z^2}{\sqrt[3]{1+z^3}} dz }
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 (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] &= \frac{1}{2}({1+z^{3}})^\frac{2}{3} + C \end{align} }