Integral of $$$\frac{e^{4 x}}{2}$$$

Find $$$\int \frac{e^{4 x}}{2}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = e^{4 x}$$$:

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

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

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

Therefore,

$$\frac{{\color{red}{\int{e^{4 x} d x}}}}{2} = \frac{{\color{red}{\int{\frac{e^{u}}{4} d u}}}}{2}$$

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

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

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 $$$u=4 x$$$:

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

Therefore,

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

Add the constant of integration:

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

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