Integral of $$$\frac{1}{a y^{4}}$$$ with respect to $$$y$$$

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

Apply the constant multiple rule $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ with $$$c=\frac{1}{a}$$$ and $$$f{\left(y \right)} = \frac{1}{y^{4}}$$$:

$${\color{red}{\int{\frac{1}{a y^{4}} d y}}} = {\color{red}{\frac{\int{\frac{1}{y^{4}} d y}}{a}}}$$

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

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

Therefore,

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

Add the constant of integration:

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

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