Integral of $$$e^{- \frac{5 x}{6}}$$$

Find $$$\int e^{- \frac{5 x}{6}}\, dx$$$.

Let $$$u=- \frac{5 x}{6}$$$.

Then $$$du=\left(- \frac{5 x}{6}\right)^{\prime }dx = - \frac{5 dx}{6}$$$ (steps can be seen »), and we have that $$$dx = - \frac{6 du}{5}$$$.

Therefore,

$${\color{red}{\int{e^{- \frac{5 x}{6}} d x}}} = {\color{red}{\int{\left(- \frac{6 e^{u}}{5}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{6}{5}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- \frac{6 e^{u}}{5}\right)d u}}} = {\color{red}{\left(- \frac{6 \int{e^{u} d u}}{5}\right)}}$$

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

$$- \frac{6 {\color{red}{\int{e^{u} d u}}}}{5} = - \frac{6 {\color{red}{e^{u}}}}{5}$$

Recall that $$$u=- \frac{5 x}{6}$$$:

$$- \frac{6 e^{{\color{red}{u}}}}{5} = - \frac{6 e^{{\color{red}{\left(- \frac{5 x}{6}\right)}}}}{5}$$

Therefore,

$$\int{e^{- \frac{5 x}{6}} d x} = - \frac{6 e^{- \frac{5 x}{6}}}{5}$$

Add the constant of integration:

$$\int{e^{- \frac{5 x}{6}} d x} = - \frac{6 e^{- \frac{5 x}{6}}}{5}+C$$

$$$\int e^{- \frac{5 x}{6}}\, dx = - \frac{6 e^{- \frac{5 x}{6}}}{5} + C$$$A