Factor out the constant using $\int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x$.
$$\frac{\int x^{3}\mathrm{d}x}{6}$$
Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{3}\mathrm{d}x$ with $\frac{x^{4}}{4}$.
$$\frac{x^{4}}{24}$$
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.
