Integral of $$$\frac{125}{6 s^{2}}$$$

Find $$$\int \frac{125}{6 s^{2}}\, ds$$$.

Apply the constant multiple rule $$$\int c f{\left(s \right)}\, ds = c \int f{\left(s \right)}\, ds$$$ with $$$c=\frac{125}{6}$$$ and $$$f{\left(s \right)} = \frac{1}{s^{2}}$$$:

$${\color{red}{\int{\frac{125}{6 s^{2}} d s}}} = {\color{red}{\left(\frac{125 \int{\frac{1}{s^{2}} d s}}{6}\right)}}$$

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

$$\frac{125 {\color{red}{\int{\frac{1}{s^{2}} d s}}}}{6}=\frac{125 {\color{red}{\int{s^{-2} d s}}}}{6}=\frac{125 {\color{red}{\frac{s^{-2 + 1}}{-2 + 1}}}}{6}=\frac{125 {\color{red}{\left(- s^{-1}\right)}}}{6}=\frac{125 {\color{red}{\left(- \frac{1}{s}\right)}}}{6}$$

Therefore,

$$\int{\frac{125}{6 s^{2}} d s} = - \frac{125}{6 s}$$

Add the constant of integration:

$$\int{\frac{125}{6 s^{2}} d s} = - \frac{125}{6 s}+C$$

$$$\int \frac{125}{6 s^{2}}\, ds = - \frac{125}{6 s} + C$$$A