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

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

Integre termo a termo:

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

A integral da função exponencial é $$$\int{e^{x} d x} = e^{x}$$$:

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

Seja $$$u=- x$$$.

Então $$$du=\left(- x\right)^{\prime }dx = - dx$$$ (veja os passos »), e obtemos $$$dx = - du$$$.

Assim,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=-1$$$ e $$$f{\left(u \right)} = e^{u}$$$:

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

A integral da função exponencial é $$$\int{e^{u} d u} = e^{u}$$$:

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

Recorde que $$$u=- x$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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