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

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

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

Therefore,

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

Add the constant of integration:

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

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