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

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

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

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

The integral can be rewritten as

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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