Integral of $$$e^{\frac{x}{a}}$$$ with respect to $$$x$$$

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

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

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

The integral becomes

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

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

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

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

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

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

$$a e^{{\color{red}{u}}} = a e^{{\color{red}{\frac{x}{a}}}}$$

Therefore,

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

Add the constant of integration:

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

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