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

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

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

So,

$${\color{red}{\int{e^{8 x} d x}}} = {\color{red}{\int{\frac{e^{u}}{8} d u}}}$$

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 $$$f{\left(u \right)} = e^{u}$$$:

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

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{8} = \frac{{\color{red}{e^{u}}}}{8}$$

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

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

Therefore,

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

Add the constant of integration:

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

Answer: $$$\int{e^{8 x} d x}=\frac{e^{8 x}}{8}+C$$$