5.5 The Substitution Rule/11: Difference between revisions
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\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] | \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 21:17, 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/de5442857b1d4d63fdfd3f9a74b2e0747dd42917)