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| \begin{align} | | \begin{align} |
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| \int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx &= \int \frac{1}{\sqrt{3ax+bx^3}}(a+bx^2)\;dx = \frac{1}{3}\int \frac{1}{\sqrt{u}}du = \int\left(\frac{1}{x}dx\right)\sin{(\ln{(x)})} \\[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] | | &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] |
Revision as of 23:23, 13 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]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/8536d7a03dc02514780eccd12207fdffdd2ad70c)