Integral de $$$e^{x} - e^{- 2 x}$$$

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

Integre termo a termo:

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

Seja $$$u=- 2 x$$$.

Então $$$du=\left(- 2 x\right)^{\prime }dx = - 2 dx$$$ (veja os passos »), e obtemos $$$dx = - \frac{du}{2}$$$.

Assim,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=- \frac{1}{2}$$$ e $$$f{\left(u \right)} = e^{u}$$$:

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

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

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

Recorde que $$$u=- 2 x$$$:

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

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

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

Portanto,

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

Adicione a constante de integração:

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

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