Integral of $$$\frac{9}{109 u^{2}}$$$

Find $$$\int \frac{9}{109 u^{2}}\, du$$$.

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{9}{109}$$$ and $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$:

$${\color{red}{\int{\frac{9}{109 u^{2}} d u}}} = {\color{red}{\left(\frac{9 \int{\frac{1}{u^{2}} d u}}{109}\right)}}$$

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

$$\frac{9 {\color{red}{\int{\frac{1}{u^{2}} d u}}}}{109}=\frac{9 {\color{red}{\int{u^{-2} d u}}}}{109}=\frac{9 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{109}=\frac{9 {\color{red}{\left(- u^{-1}\right)}}}{109}=\frac{9 {\color{red}{\left(- \frac{1}{u}\right)}}}{109}$$

Therefore,

$$\int{\frac{9}{109 u^{2}} d u} = - \frac{9}{109 u}$$

Add the constant of integration:

$$\int{\frac{9}{109 u^{2}} d u} = - \frac{9}{109 u}+C$$

$$$\int \frac{9}{109 u^{2}}\, du = - \frac{9}{109 u} + C$$$A