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

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

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

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

So,

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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