Answer
$\frac{20}{3}\sqrt{1+x^3}+C$
Work Step by Step
$\frac{dy}{dx}=\frac{10x^2}{\sqrt{1+x^3}}$
$\int\frac{10x^2}{\sqrt{1+x^3}}dx$
$u=1+x^3\hspace{8mm}du=3x^2dx\hspace{8mm}\frac{du}{3x^2}=dx$
Therefore
$\int\frac{10x^2}{\sqrt{1+x^3}}dx=\int\frac{10x^2}{\sqrt{u}}\frac{1}{3x^2}du=\frac{10}{3}\int\frac{1}{\sqrt{u}}du=\frac{10}{3}\int u^{-\frac{1}{2}}du$
$=\frac{10}{3}(2u^{\frac{1}{2}})+C=\frac{20}{3}\sqrt{1+x^3}+C$
