Integral of $$$\frac{1}{116 k^{2}}$$$

Find $$$\int \frac{1}{116 k^{2}}\, dk$$$.

Apply the constant multiple rule $$$\int c f{\left(k \right)}\, dk = c \int f{\left(k \right)}\, dk$$$ with $$$c=\frac{1}{116}$$$ and $$$f{\left(k \right)} = \frac{1}{k^{2}}$$$:

$${\color{red}{\int{\frac{1}{116 k^{2}} d k}}} = {\color{red}{\left(\frac{\int{\frac{1}{k^{2}} d k}}{116}\right)}}$$

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

$$\frac{{\color{red}{\int{\frac{1}{k^{2}} d k}}}}{116}=\frac{{\color{red}{\int{k^{-2} d k}}}}{116}=\frac{{\color{red}{\frac{k^{-2 + 1}}{-2 + 1}}}}{116}=\frac{{\color{red}{\left(- k^{-1}\right)}}}{116}=\frac{{\color{red}{\left(- \frac{1}{k}\right)}}}{116}$$

Therefore,

$$\int{\frac{1}{116 k^{2}} d k} = - \frac{1}{116 k}$$

Add the constant of integration:

$$\int{\frac{1}{116 k^{2}} d k} = - \frac{1}{116 k}+C$$

$$$\int \frac{1}{116 k^{2}}\, dk = - \frac{1}{116 k} + C$$$A