Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{14}\mathrm{d}x$ with $\frac{x^{15}}{15}$. Multiply $0.3$ times $\frac{x^{15}}{15}$.
Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{13}\mathrm{d}x$ with $\frac{x^{14}}{14}$. Multiply $0.3$ times $\frac{x^{14}}{14}$.
$$\frac{x^{15}}{50}+\frac{3x^{14}}{140}$$
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.
