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

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

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 $$$f{\left(x \right)} = e^{2 x}$$$:

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

Let $$$u=2 x$$$.

Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

The integral becomes

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

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

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

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 $$$u=2 x$$$:

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

Therefore,

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

Add the constant of integration:

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

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