5.5 The Substitution Rule/17

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

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

${\begin{aligned}\int {\frac {a+bx^{2}}{\sqrt {3ax+bx^{3}}}}dx&=\int {\frac {1}{\sqrt {3ax+bx^{3}}}}(a+bx^{2})\;dx={\frac {1}{3}}\int {\frac {1}{\sqrt {u}}}du=\int \left({\frac {1}{x}}dx\right)\sin {(\ln {(x)})}\\[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}$