Integral de $$$e^{x} + e^{- x}$$$

Halla $$$\int \left(e^{x} + e^{- x}\right)\, dx$$$.

Integra término a término:

$${\color{red}{\int{\left(e^{x} + e^{- x}\right)d x}}} = {\color{red}{\left(\int{e^{- x} d x} + \int{e^{x} d x}\right)}}$$

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

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

Sea $$$u=- x$$$.

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

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = e^{u}$$$:

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

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

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

Recordemos que $$$u=- x$$$:

$$e^{x} - e^{{\color{red}{u}}} = e^{x} - e^{{\color{red}{\left(- x\right)}}}$$

Por lo tanto,

$$\int{\left(e^{x} + e^{- x}\right)d x} = e^{x} - e^{- x}$$

Simplificar:

$$\int{\left(e^{x} + e^{- x}\right)d x} = 2 \sinh{\left(x \right)}$$

Añade la constante de integración:

$$\int{\left(e^{x} + e^{- x}\right)d x} = 2 \sinh{\left(x \right)}+C$$

$$$\int \left(e^{x} + e^{- x}\right)\, dx = 2 \sinh{\left(x \right)} + C$$$A