Integral of $$$8 e^{- 8 x}$$$

Find $$$\int 8 e^{- 8 x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=8$$$ and $$$f{\left(x \right)} = e^{- 8 x}$$$:

$${\color{red}{\int{8 e^{- 8 x} d x}}} = {\color{red}{\left(8 \int{e^{- 8 x} d x}\right)}}$$

Let $$$u=- 8 x$$$.

Then $$$du=\left(- 8 x\right)^{\prime }dx = - 8 dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{8}$$$.

The integral becomes

$$8 {\color{red}{\int{e^{- 8 x} d x}}} = 8 {\color{red}{\int{\left(- \frac{e^{u}}{8}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{8}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$8 {\color{red}{\int{\left(- \frac{e^{u}}{8}\right)d u}}} = 8 {\color{red}{\left(- \frac{\int{e^{u} d u}}{8}\right)}}$$

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

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

Recall that $$$u=- 8 x$$$:

$$- e^{{\color{red}{u}}} = - e^{{\color{red}{\left(- 8 x\right)}}}$$

Therefore,

$$\int{8 e^{- 8 x} d x} = - e^{- 8 x}$$

Add the constant of integration:

$$\int{8 e^{- 8 x} d x} = - e^{- 8 x}+C$$

$$$\int 8 e^{- 8 x}\, dx = - e^{- 8 x} + C$$$A