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

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{1}{u^{2}}\, du = - \frac{1}{u} + C$$$A