Integral de $$$- b^{- x} + a^{- x}$$$ em relação a $$$x$$$

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

Integre termo a termo:

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

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

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

A integral torna-se

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

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

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

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=a$$$:

$$- \int{b^{- x} d x} - {\color{red}{\int{a^{u} d u}}} = - \int{b^{- x} d x} - {\color{red}{\frac{a^{u}}{\ln{\left(a \right)}}}}$$

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

$$- \int{b^{- x} d x} - \frac{a^{{\color{red}{u}}}}{\ln{\left(a \right)}} = - \int{b^{- x} d x} - \frac{a^{{\color{red}{\left(- x\right)}}}}{\ln{\left(a \right)}}$$

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

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

Logo,

$$- {\color{red}{\int{b^{- x} d x}}} - \frac{a^{- x}}{\ln{\left(a \right)}} = - {\color{red}{\int{\left(- b^{u}\right)d u}}} - \frac{a^{- x}}{\ln{\left(a \right)}}$$

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

$$- {\color{red}{\int{\left(- b^{u}\right)d u}}} - \frac{a^{- x}}{\ln{\left(a \right)}} = - {\color{red}{\left(- \int{b^{u} d u}\right)}} - \frac{a^{- x}}{\ln{\left(a \right)}}$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=b$$$:

$${\color{red}{\int{b^{u} d u}}} - \frac{a^{- x}}{\ln{\left(a \right)}} = {\color{red}{\frac{b^{u}}{\ln{\left(b \right)}}}} - \frac{a^{- x}}{\ln{\left(a \right)}}$$

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

$$\frac{b^{{\color{red}{u}}}}{\ln{\left(b \right)}} - \frac{a^{- x}}{\ln{\left(a \right)}} = \frac{b^{{\color{red}{\left(- x\right)}}}}{\ln{\left(b \right)}} - \frac{a^{- x}}{\ln{\left(a \right)}}$$

Portanto,

$$\int{\left(- b^{- x} + a^{- x}\right)d x} = \frac{b^{- x}}{\ln{\left(b \right)}} - \frac{a^{- x}}{\ln{\left(a \right)}}$$

Adicione a constante de integração:

$$\int{\left(- b^{- x} + a^{- x}\right)d x} = \frac{b^{- x}}{\ln{\left(b \right)}} - \frac{a^{- x}}{\ln{\left(a \right)}}+C$$

$$$\int \left(- b^{- x} + a^{- x}\right)\, dx = \left(\frac{b^{- x}}{\ln\left(b\right)} - \frac{a^{- x}}{\ln\left(a\right)}\right) + C$$$A