Integral of $$$e^{- 4 x}$$$

Let $$$u=- 4 x$$$.

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

The integral becomes

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{4}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

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

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

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

Recall that $$$u=- 4 x$$$:

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

Therefore,

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

Add the constant of integration:

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

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