Integral of $$$\frac{1}{y^{4}}$$$

Find $$$\int \frac{1}{y^{4}}\, dy$$$.

Apply the power rule $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:

$${\color{red}{\int{\frac{1}{y^{4}} d y}}}={\color{red}{\int{y^{-4} d y}}}={\color{red}{\frac{y^{-4 + 1}}{-4 + 1}}}={\color{red}{\left(- \frac{y^{-3}}{3}\right)}}={\color{red}{\left(- \frac{1}{3 y^{3}}\right)}}$$

Therefore,

$$\int{\frac{1}{y^{4}} d y} = - \frac{1}{3 y^{3}}$$

Add the constant of integration:

$$\int{\frac{1}{y^{4}} d y} = - \frac{1}{3 y^{3}}+C$$

$$$\int \frac{1}{y^{4}}\, dy = - \frac{1}{3 y^{3}} + C$$$A