Rewrite $\frac{1}{\sqrt[3]{x}}$ as $x^{-\frac{1}{3}}$. Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{-\frac{1}{3}}\mathrm{d}x$ with $\frac{x^{\frac{2}{3}}}{\frac{2}{3}}$. Simplify.
$$\frac{3x^{\frac{2}{3}}}{2}$$
If $F\left(x\right)$ is an antiderivative of $f\left(x\right)$, then the set of all antiderivatives of $f\left(x\right)$ is given by $F\left(x\right)+C$. Therefore, add the constant of integration $C\in \mathrm{R}$ to the result.
