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

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

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

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

The integral becomes

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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