Integral of $$$\frac{1}{y^{3}}$$$

Find $$$\int \frac{1}{y^{3}}\, dy$$$.

Apply the power rule $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

$${\color{red}{\int{\frac{1}{y^{3}} d y}}}={\color{red}{\int{y^{-3} d y}}}={\color{red}{\frac{y^{-3 + 1}}{-3 + 1}}}={\color{red}{\left(- \frac{y^{-2}}{2}\right)}}={\color{red}{\left(- \frac{1}{2 y^{2}}\right)}}$$

Therefore,

$$\int{\frac{1}{y^{3}} d y} = - \frac{1}{2 y^{2}}$$

Add the constant of integration:

$$\int{\frac{1}{y^{3}} d y} = - \frac{1}{2 y^{2}}+C$$

$$$\int \frac{1}{y^{3}}\, dy = - \frac{1}{2 y^{2}} + C$$$A