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

Encontre $$$\int \frac{1}{x^{2}}\, dx$$$.

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

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

Portanto,

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

Adicione a constante de integração:

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

$$$\int \frac{1}{x^{2}}\, dx = - \frac{1}{x} + C$$$A