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

Find $$$\int \frac{1}{z^{2}}\, dz$$$.

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{1}{z^{2}}\, dz = - \frac{1}{z} + C$$$A