5.5 The Substitution Rule/11: Difference between revisions
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\int (x+1)\sqrt{2x+x^{2}}dx | \int (x+1)\sqrt{2x+x^{2}}dx | ||
</math> | </math> | ||
<math> | <math> | ||
| Line 12: | Line 13: | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
\int (x+1)\sqrt{2x+x^{2}}dx &= \frac{1}{2}\int\sqrt{u}du | \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] | ||
&= \frac{1}{2}({\frac{2u^\frac{3}{2}}{3}}) + C \\[2ex] | |||
&= \frac{1}{3}(u^{\frac{3}{2}}) + C \\[2ex] | |||
&= \frac{1}{3}(2x+x^{2})^{\frac{3}{2}} + C \\[2ex] | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 21:28, 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]&={\frac {1}{2}}({\frac {2u^{\frac {3}{2}}}{3}})+C\\[2ex]&={\frac {1}{3}}(u^{\frac {3}{2}})+C\\[2ex]&={\frac {1}{3}}(2x+x^{2})^{\frac {3}{2}}+C\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/64df293c2132ddffb8fda9e21d7f148a4b52200a)