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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

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

Therefore,

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

Add the constant of integration:

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

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