5.5 The Substitution Rule/17: Difference between revisions
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\begin{align} | \begin{align} | ||
\int \frac{a+bx^2}{\sqrt{3ax+bx^3}}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{a+bx^2}{\sqrt{3ax+bx^3}}dx = \frac{1}{3}\int \frac{1}{\sqrt{u}}du = \int\left(\frac{1}{x}dx\right)\sin{(\ln{(x)})} \\[2ex] | ||
&= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | ||
Revision as of 23:12, 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 {a+bx^{2}}{\sqrt {3ax+bx^{3}}}}dx={\frac {1}{3}}\int {\frac {1}{\sqrt {u}}}du=\int \left({\frac {1}{x}}dx\right)\sin {(\ln {(x)})}\\[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/cd510e8098d470ae4d69d5464eb04518be914c68)