Integral of $$$e^{- \frac{3 x}{4}}$$$

Find $$$\int e^{- \frac{3 x}{4}}\, dx$$$.

Let $$$u=- \frac{3 x}{4}$$$.

Then $$$du=\left(- \frac{3 x}{4}\right)^{\prime }dx = - \frac{3 dx}{4}$$$ (steps can be seen »), and we have that $$$dx = - \frac{4 du}{3}$$$.

Therefore,

$${\color{red}{\int{e^{- \frac{3 x}{4}} d x}}} = {\color{red}{\int{\left(- \frac{4 e^{u}}{3}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{4}{3}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- \frac{4 e^{u}}{3}\right)d u}}} = {\color{red}{\left(- \frac{4 \int{e^{u} d u}}{3}\right)}}$$

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

$$- \frac{4 {\color{red}{\int{e^{u} d u}}}}{3} = - \frac{4 {\color{red}{e^{u}}}}{3}$$

Recall that $$$u=- \frac{3 x}{4}$$$:

$$- \frac{4 e^{{\color{red}{u}}}}{3} = - \frac{4 e^{{\color{red}{\left(- \frac{3 x}{4}\right)}}}}{3}$$

Therefore,

$$\int{e^{- \frac{3 x}{4}} d x} = - \frac{4 e^{- \frac{3 x}{4}}}{3}$$

Add the constant of integration:

$$\int{e^{- \frac{3 x}{4}} d x} = - \frac{4 e^{- \frac{3 x}{4}}}{3}+C$$

$$$\int e^{- \frac{3 x}{4}}\, dx = - \frac{4 e^{- \frac{3 x}{4}}}{3} + C$$$A