Answer
$\dfrac{2}{3} \sin^{-1} (x^{3/2})+C$
Work Step by Step
Consider $u=x^{3/2} \implies dx=\dfrac{2 du}{3\sqrt u}$
Now, the given integral can be written as:
$\dfrac{2}{3}\dfrac{1}{\sqrt{1-u^2}} du=\dfrac{2}{3} \sin^{-1} u+C$
or, $=\dfrac{2}{3} \sin^{-1} (x^{3/2})+C$
