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 $2$ times $\frac{x^{4}}{4}$.
Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x\mathrm{d}x$ with $\frac{x^{2}}{2}$. Multiply $-5$ times $\frac{x^{2}}{2}$.
Find the integral of $7$ using the table of common integrals rule $\int a\mathrm{d}x=ax$.
$$\frac{x^{4}}{2}-\frac{5x^{2}}{2}+7x$$
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.
