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

Find $$$\int \frac{1}{p^{2}}\, dp$$$.

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{1}{p^{2}}\, dp = - \frac{1}{p} + C$$$A