$\begin{align}  & \int{\sqrt{{{x}^{3}}}}dx=\int{{{x}^{\tfrac{1}{3}}}dx}=\frac{{{x}^{\tfrac{1}{3}+1}}}{\tfrac{1}{3}+1}+C=\frac{{{x}^{\tfrac{4}{3}}}}{\tfrac{4}{3}}+C \\ & \text{ }\int{\sqrt{{{x}^{3}}}}dx=\frac{3}{4}{{x}^{\tfrac{4}{3}}}+C \\\end{align}$