5.5 The Substitution Rule/11: Difference between revisions
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(Created page with "<math> \int (x+1)\sqrt{2x+x^{2}}dx </math> <math> \begin{align} u &=2x+x^{2} \\[2ex] du &=2+2x dx \\[2ex] \frac{1}{2}du &=x+1 \end{align} </math>") |
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du &=2+2x dx \\[2ex] | du &=2+2x dx \\[2ex] | ||
\frac{1}{2}du &=x+1 | \frac{1}{2}du &=x+1 | ||
\end{align} | |||
</math> | |||
<math> | |||
\begin{align} | |||
\int (x+1)\sqrt{2x+x^{2}}dx &= \frac{1}{2}\int\sqrt{u}du &= \frac{1}{2}\int u^{\frac{1}{2}}du \\[2ex] | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 20:09, 22 September 2022
![{\displaystyle {\begin{aligned}u&=2x+x^{2}\\[2ex]du&=2+2xdx\\[2ex]{\frac {1}{2}}du&=x+1\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/457f3b114a8832535d70a7a359fc2dbe034c038b)
![{\displaystyle {\begin{aligned}\int (x+1){\sqrt {2x+x^{2}}}dx&={\frac {1}{2}}\int {\sqrt {u}}du&={\frac {1}{2}}\int u^{\frac {1}{2}}du\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ea090c66a0778f82b8b29827b96a136217d02404)