Integral de $$$e^{\frac{y}{x}}$$$ con respecto a $$$y$$$

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

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

Entonces $$$du=\left(\frac{y}{x}\right)^{\prime }dy = \frac{dy}{x}$$$ (los pasos pueden verse »), y obtenemos que $$$dy = x du$$$.

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=x$$$ y $$$f{\left(u \right)} = e^{u}$$$:

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

La integral de la función exponencial es $$$\int{e^{u} d u} = e^{u}$$$:

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

Recordemos que $$$u=\frac{y}{x}$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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