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

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

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