Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{4}\mathrm{d}x$ with $\frac{x^{5}}{5}$. Multiply $3$ times $\frac{x^{5}}{5}$.
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 $5$ times $\frac{x^{3}}{3}$.
Find the integral of $-2$ using the table of common integrals rule $\int a\mathrm{d}x=ax$.
$$\frac{3x^{5}}{5}+\frac{5x^{3}}{3}-2x$$
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.
