Integral of $$$8 e^{4 x}$$$

Find $$$\int 8 e^{4 x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=8$$$ and $$$f{\left(x \right)} = e^{4 x}$$$:

$${\color{red}{\int{8 e^{4 x} d x}}} = {\color{red}{\left(8 \int{e^{4 x} d x}\right)}}$$

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

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

So,

$$8 {\color{red}{\int{e^{4 x} d x}}} = 8 {\color{red}{\int{\frac{e^{u}}{4} d u}}}$$

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

$$8 {\color{red}{\int{\frac{e^{u}}{4} d u}}} = 8 {\color{red}{\left(\frac{\int{e^{u} d u}}{4}\right)}}$$

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

$$2 {\color{red}{\int{e^{u} d u}}} = 2 {\color{red}{e^{u}}}$$

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

$$2 e^{{\color{red}{u}}} = 2 e^{{\color{red}{\left(4 x\right)}}}$$

Therefore,

$$\int{8 e^{4 x} d x} = 2 e^{4 x}$$

Add the constant of integration:

$$\int{8 e^{4 x} d x} = 2 e^{4 x}+C$$

$$$\int 8 e^{4 x}\, dx = 2 e^{4 x} + C$$$A