Integral of $$$\frac{1}{4 y^{3}}$$$

Find $$$\int \frac{1}{4 y^{3}}\, dy$$$.

Apply the constant multiple rule $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(y \right)} = \frac{1}{y^{3}}$$$:

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

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

$$\frac{{\color{red}{\int{\frac{1}{y^{3}} d y}}}}{4}=\frac{{\color{red}{\int{y^{-3} d y}}}}{4}=\frac{{\color{red}{\frac{y^{-3 + 1}}{-3 + 1}}}}{4}=\frac{{\color{red}{\left(- \frac{y^{-2}}{2}\right)}}}{4}=\frac{{\color{red}{\left(- \frac{1}{2 y^{2}}\right)}}}{4}$$

Therefore,

$$\int{\frac{1}{4 y^{3}} d y} = - \frac{1}{8 y^{2}}$$

Add the constant of integration:

$$\int{\frac{1}{4 y^{3}} d y} = - \frac{1}{8 y^{2}}+C$$

$$$\int \frac{1}{4 y^{3}}\, dy = - \frac{1}{8 y^{2}} + C$$$A