Integral de $$$\frac{1}{r^{2}}$$$

Encontre $$$\int \frac{1}{r^{2}}\, dr$$$.

Aplique a regra da potência $$$\int r^{n}\, dr = \frac{r^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=-2$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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