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

Find $$$\int \frac{1}{k^{2}}\, dk$$$.

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{1}{k^{2}}\, dk = - \frac{1}{k} + C$$$A