Integral of $$$\frac{1}{c^{2} y^{2}}$$$ with respect to $$$y$$$

Find $$$\int \frac{1}{c^{2} y^{2}}\, dy$$$.

Apply the constant multiple rule $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ with $$$c=\frac{1}{c^{2}}$$$ and $$$f{\left(y \right)} = \frac{1}{y^{2}}$$$:

$${\color{red}{\int{\frac{1}{c^{2} y^{2}} d y}}} = {\color{red}{\frac{\int{\frac{1}{y^{2}} d y}}{c^{2}}}}$$

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{1}{c^{2} y^{2}}\, dy = - \frac{1}{c^{2} y} + C$$$A