Integral of $$$e^{\frac{2 y}{x}}$$$ with respect to $$$y$$$

Find $$$\int e^{\frac{2 y}{x}}\, dy$$$.

Let $$$u=\frac{2 y}{x}$$$.

Then $$$du=\left(\frac{2 y}{x}\right)^{\prime }dy = \frac{2}{x} dy$$$ (steps can be seen »), and we have that $$$dy = \frac{x du}{2}$$$.

So,

$${\color{red}{\int{e^{\frac{2 y}{x}} d y}}} = {\color{red}{\int{\frac{x e^{u}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{x}{2}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\frac{x e^{u}}{2} d u}}} = {\color{red}{\left(\frac{x \int{e^{u} d u}}{2}\right)}}$$

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

$$\frac{x {\color{red}{\int{e^{u} d u}}}}{2} = \frac{x {\color{red}{e^{u}}}}{2}$$

Recall that $$$u=\frac{2 y}{x}$$$:

$$\frac{x e^{{\color{red}{u}}}}{2} = \frac{x e^{{\color{red}{\left(\frac{2 y}{x}\right)}}}}{2}$$

Therefore,

$$\int{e^{\frac{2 y}{x}} d y} = \frac{x e^{\frac{2 y}{x}}}{2}$$

Add the constant of integration:

$$\int{e^{\frac{2 y}{x}} d y} = \frac{x e^{\frac{2 y}{x}}}{2}+C$$

$$$\int e^{\frac{2 y}{x}}\, dy = \frac{x e^{\frac{2 y}{x}}}{2} + C$$$A