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

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

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

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

Therefore,

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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