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 $\frac{x^{4}}{4}$. Multiply $-1$ times $\frac{x^{4}}{4}$.
$$x-\frac{x^{4}}{4}-3\int x^{2}\mathrm{d}x$$
Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{2}\mathrm{d}x$ with $\frac{x^{3}}{3}$. Multiply $-3$ times $\frac{x^{3}}{3}$.
$$x-\frac{x^{4}}{4}-x^{3}$$
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.
