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

Let $$$u=\frac{x}{3}$$$.

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

So,

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

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

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

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

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

Recall that $$$u=\frac{x}{3}$$$:

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

Therefore,

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

Add the constant of integration:

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

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