Integral of $$$\frac{e^{3 x}}{3}$$$

Find $$$\int \frac{e^{3 x}}{3}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(x \right)} = e^{3 x}$$$:

$${\color{red}{\int{\frac{e^{3 x}}{3} d x}}} = {\color{red}{\left(\frac{\int{e^{3 x} d x}}{3}\right)}}$$

Let $$$u=3 x$$$.

Then $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

Thus,

$$\frac{{\color{red}{\int{e^{3 x} d x}}}}{3} = \frac{{\color{red}{\int{\frac{e^{u}}{3} d u}}}}{3}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$\frac{{\color{red}{\int{\frac{e^{u}}{3} d u}}}}{3} = \frac{{\color{red}{\left(\frac{\int{e^{u} d u}}{3}\right)}}}{3}$$

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{9} = \frac{{\color{red}{e^{u}}}}{9}$$

Recall that $$$u=3 x$$$:

$$\frac{e^{{\color{red}{u}}}}{9} = \frac{e^{{\color{red}{\left(3 x\right)}}}}{9}$$

Therefore,

$$\int{\frac{e^{3 x}}{3} d x} = \frac{e^{3 x}}{9}$$

Add the constant of integration:

$$\int{\frac{e^{3 x}}{3} d x} = \frac{e^{3 x}}{9}+C$$

$$$\int \frac{e^{3 x}}{3}\, dx = \frac{e^{3 x}}{9} + C$$$A