Integral of $$$\frac{1}{a^{2} t^{2}}$$$ with respect to $$$t$$$

Find $$$\int \frac{1}{a^{2} t^{2}}\, dt$$$.

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

$${\color{red}{\int{\frac{1}{a^{2} t^{2}} d t}}} = {\color{red}{\frac{\int{\frac{1}{t^{2}} d t}}{a^{2}}}}$$

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{1}{a^{2} t^{2}}\, dt = - \frac{1}{a^{2} t} + C$$$A