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

Find $$$\int \frac{1}{v^{2}}\, dv$$$.

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{1}{v^{2}}\, dv = - \frac{1}{v} + C$$$A