5.5 The Substitution Rule/17

$\int {\frac {a+bx^{2}}{\sqrt {3ax+bx^{3}}}}dx$

${\begin{aligned}u&=3ax+bx^{3}\\[2ex]du&=(3a+3bx^{2})dx\\[2ex]{\frac {1}{3}}du&=(a+bx^{2})dx\\[2ex]\end{aligned}}$

Failed to parse (unknown function "\rightdx"): {\displaystyle \begin{align} \int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx &= \int \frac{1}{\sqrt{3ax+bx^3}}\left(a+bx^2)\rightdx = \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{align} }