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| \begin{align} | | \begin{align} |
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| \int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx &= \frac{1}{3}\int \frac{1}{\sqrt{u}}du = \frac{1}{3}\int \{u^{-1/2}du [2ex] | | \int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx &= \int \frac{1}{\sqrt{3ax+bx^3}}(a+bx^2)\;dx = \int \frac{1}{\sqrt{3ax+bx^3}}((a+bx^2)\;dx)\ \\[2ex] |
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| &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | | |
| &= -\cos{(u)} + C \\[2ex] | | &= \frac{1}{3}\int \frac{1}{\sqrt{u}}(du) = \frac{1}{3}\int u^{-1/2} du \\[2ex] |
| &= -\cos{(\ln{(x)})} + C | | &= \frac{1}{3}\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C \\[2ex] |
| | &= \frac{2}{3}(3ax+bx^3)^{1/2} + C \\[2ex] |
| | &= \frac{2}{3}{\sqrt{3ax+bx^3}} + C |
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| \end{align} | | \end{align} |
| </math> | | </math> |
Latest revision as of 19:18, 20 September 2022
![{\displaystyle {\begin{aligned}u&=3ax+bx^{3}\\[2ex]du&=(3a+3bx^{2})dx\\[2ex]{\frac {1}{3}}du&=(a+bx^{2})dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/02464f0e436c19b198f5cf8d28fa89ba4633df4c)
![{\displaystyle {\begin{aligned}\int {\frac {a+bx^{2}}{\sqrt {3ax+bx^{3}}}}dx&=\int {\frac {1}{\sqrt {3ax+bx^{3}}}}(a+bx^{2})\;dx=\int {\frac {1}{\sqrt {3ax+bx^{3}}}}((a+bx^{2})\;dx)\ \\[2ex]&={\frac {1}{3}}\int {\frac {1}{\sqrt {u}}}(du)={\frac {1}{3}}\int u^{-1/2}du\\[2ex]&={\frac {1}{3}}{\frac {u^{\frac {1}{2}}}{\frac {1}{2}}}+C\\[2ex]&={\frac {2}{3}}(3ax+bx^{3})^{1/2}+C\\[2ex]&={\frac {2}{3}}{\sqrt {3ax+bx^{3}}}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/df3c9c28bb88c787ac75087ea25d5e364911b25c)