Integral of $$$- \frac{e^{u}}{9}$$$

Find $$$\int \left(- \frac{e^{u}}{9}\right)\, du$$$.

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

$${\color{red}{\int{\left(- \frac{e^{u}}{9}\right)d u}}} = {\color{red}{\left(- \frac{\int{e^{u} d u}}{9}\right)}}$$

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

$$- \frac{{\color{red}{\int{e^{u} d u}}}}{9} = - \frac{{\color{red}{e^{u}}}}{9}$$

Therefore,

$$\int{\left(- \frac{e^{u}}{9}\right)d u} = - \frac{e^{u}}{9}$$

Add the constant of integration:

$$\int{\left(- \frac{e^{u}}{9}\right)d u} = - \frac{e^{u}}{9}+C$$

$$$\int \left(- \frac{e^{u}}{9}\right)\, du = - \frac{e^{u}}{9} + C$$$A