Answer
$\frac{2}{3}\sqrt{x^3+3x}+C$
Work Step by Step
$\int\frac{x^2+1}{\sqrt{x^3+3x}}dx$
Let $u=x^3+3x$. Then $du=(3x^2+3)dx$, so $(x^2+1)dx=\frac{du}{3}$.
$=\int\frac{1}{\sqrt{u}}*\frac{du}{3}$
$=\frac{1}{3}\int u^{-\frac{1}{2}} du$
$=\frac{1}{3}(\frac{u^\frac{1}{2}}{\frac{1}{2}}+C)$
$=\frac{2}{3}\sqrt{u}+C$
$=\frac{2}{3}\sqrt{x^3+3x}+C$
