Integral of $$$\frac{1}{r^{3}}$$$

Find $$$\int \frac{1}{r^{3}}\, dr$$$.

Apply the power rule $$$\int r^{n}\, dr = \frac{r^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

$${\color{red}{\int{\frac{1}{r^{3}} d r}}}={\color{red}{\int{r^{-3} d r}}}={\color{red}{\frac{r^{-3 + 1}}{-3 + 1}}}={\color{red}{\left(- \frac{r^{-2}}{2}\right)}}={\color{red}{\left(- \frac{1}{2 r^{2}}\right)}}$$

Therefore,

$$\int{\frac{1}{r^{3}} d r} = - \frac{1}{2 r^{2}}$$

Add the constant of integration:

$$\int{\frac{1}{r^{3}} d r} = - \frac{1}{2 r^{2}}+C$$

$$$\int \frac{1}{r^{3}}\, dr = - \frac{1}{2 r^{2}} + C$$$A