Answer
$\int\sqrt[3]{x^2} dx=\dfrac{3\sqrt[3]{x^5}}{5}+C$
Work Step by Step
$\int\sqrt[3]{x^2} dx=\int x^{\frac{2}{3}}dx=\dfrac{1}{\frac{2}{3}+1}\times x^{\frac{2}{3}+1}+C=\dfrac{3\sqrt[3]{x^5}}{5}+C$
By differentiating, we get the same integrand we started with.
