Answer
$\frac{2\sqrt{3ax+bx^3}}{3} +C$
Work Step by Step
Evaluate the Integral using substitution: $\int \frac{a+bx^2}{\sqrt{3ax+bx^3}}$
Substitution Rule: $\int f(g(x))g’(x)dx = \int f(u)du$
$u= 3ax+bx^3$
$du =3a + 3bx^2$
Since $du$ in the expression is equal to $(a+bx^2)$ it must be multiplied by $\frac{1}{3}$
Solve the integral in terms of $u$:
$\int \frac{1}{\sqrt u}(\frac13)du$
$\frac{1}{3}\int \frac{1}{\sqrt u}du $
$\frac{1}{3}2\sqrt u +C$
Substitute for $u$:
$\frac{2\sqrt{3ax+bx^3}}{3} +C$
