Answer
$$\int\frac{a+bx^2}{\sqrt{3ax+bx^3}}dx=\frac{2\sqrt{3ax+bx^3}}{3}+C$$
Work Step by Step
$$A=\int\frac{a+bx^2}{\sqrt{3ax+bx^3}}dx$$
Let $u=3ax+bx^3$.
Then $du=(3a+3bx^2)dx=3(a+bx^2)dx$, so $(a+bx^2)dx=\frac{1}{3}du$
Also, $\sqrt{3ax+bx^3}=\sqrt u=u^{1/2}$
Substitute into $A$, we have $$A=\frac{1}{3}\int\frac{1}{u^{1/2}}du$$ $$A=\frac{1}{3}\int u^{-1/2}du$$ $$A=\frac{1}{3}\frac{u^{1/2}}{\frac{1}{2}}+C$$ $$A=\frac{\sqrt u}{\frac{3}{2}}+C$$ $$A=\frac{2\sqrt u}{3}+C$$ $$A=\frac{2\sqrt{3ax+bx^3}}{3}+C$$
