Integral de $$$\left(e^{x} + 2\right) e^{- x}$$$

Encontre $$$\int \left(e^{x} + 2\right) e^{- x}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\left(e^{x} + 2\right) e^{- x} d x}}} = {\color{red}{\int{\left(1 + 2 e^{- x}\right)d x}}}$$

Integre termo a termo:

$${\color{red}{\int{\left(1 + 2 e^{- x}\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{2 e^{- x} d x}\right)}}$$

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=1$$$:

$$\int{2 e^{- x} d x} + {\color{red}{\int{1 d x}}} = \int{2 e^{- x} d x} + {\color{red}{x}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=2$$$ e $$$f{\left(x \right)} = e^{- x}$$$:

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

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

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

Portanto,

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

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

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

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

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

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

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

Portanto,

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

Adicione a constante de integração:

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

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