Integral de $$$7^{- \frac{1}{x}}$$$

Encontre $$$\int 7^{- \frac{1}{x}}\, dx$$$.

Alterar a base:

$${\color{red}{\int{7^{- \frac{1}{x}} d x}}} = {\color{red}{\int{e^{- \frac{\ln{\left(7 \right)}}{x}} d x}}}$$

Para a integral $$$\int{e^{- \frac{\ln{\left(7 \right)}}{x}} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=e^{- \frac{\ln{\left(7 \right)}}{x}}$$$ e $$$\operatorname{dv}=dx$$$.

Então $$$\operatorname{du}=\left(e^{- \frac{\ln{\left(7 \right)}}{x}}\right)^{\prime }dx=\frac{e^{- \frac{\ln{\left(7 \right)}}{x}} \ln{\left(7 \right)}}{x^{2}} dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{1 d x}=x$$$ (os passos podem ser vistos »).

A integral pode ser reescrita como

$${\color{red}{\int{e^{- \frac{\ln{\left(7 \right)}}{x}} d x}}}={\color{red}{\left(e^{- \frac{\ln{\left(7 \right)}}{x}} \cdot x-\int{x \cdot \frac{e^{- \frac{\ln{\left(7 \right)}}{x}} \ln{\left(7 \right)}}{x^{2}} d x}\right)}}={\color{red}{\left(x e^{- \frac{\ln{\left(7 \right)}}{x}} - \int{\frac{e^{- \frac{\ln{\left(7 \right)}}{x}} \ln{\left(7 \right)}}{x} d x}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\ln{\left(7 \right)}$$$ e $$$f{\left(x \right)} = \frac{e^{- \frac{\ln{\left(7 \right)}}{x}}}{x}$$$:

$$x e^{- \frac{\ln{\left(7 \right)}}{x}} - {\color{red}{\int{\frac{e^{- \frac{\ln{\left(7 \right)}}{x}} \ln{\left(7 \right)}}{x} d x}}} = x e^{- \frac{\ln{\left(7 \right)}}{x}} - {\color{red}{\ln{\left(7 \right)} \int{\frac{e^{- \frac{\ln{\left(7 \right)}}{x}}}{x} d x}}}$$

Seja $$$u=- \frac{\ln{\left(7 \right)}}{x}$$$.

Então $$$du=\left(- \frac{\ln{\left(7 \right)}}{x}\right)^{\prime }dx = \frac{\ln{\left(7 \right)}}{x^{2}} dx$$$ (veja os passos »), e obtemos $$$\frac{dx}{x^{2}} = \frac{du}{\ln{\left(7 \right)}}$$$.

Logo,

$$x e^{- \frac{\ln{\left(7 \right)}}{x}} - \ln{\left(7 \right)} {\color{red}{\int{\frac{e^{- \frac{\ln{\left(7 \right)}}{x}}}{x} d x}}} = x e^{- \frac{\ln{\left(7 \right)}}{x}} - \ln{\left(7 \right)} {\color{red}{\int{\left(- \frac{e^{u}}{u}\right)d u}}}$$

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

$$x e^{- \frac{\ln{\left(7 \right)}}{x}} - \ln{\left(7 \right)} {\color{red}{\int{\left(- \frac{e^{u}}{u}\right)d u}}} = x e^{- \frac{\ln{\left(7 \right)}}{x}} - \ln{\left(7 \right)} {\color{red}{\left(- \int{\frac{e^{u}}{u} d u}\right)}}$$

Esta integral (Integral Exponencial) não possui forma fechada:

$$x e^{- \frac{\ln{\left(7 \right)}}{x}} + \ln{\left(7 \right)} {\color{red}{\int{\frac{e^{u}}{u} d u}}} = x e^{- \frac{\ln{\left(7 \right)}}{x}} + \ln{\left(7 \right)} {\color{red}{\operatorname{Ei}{\left(u \right)}}}$$

Recorde que $$$u=- \frac{\ln{\left(7 \right)}}{x}$$$:

$$x e^{- \frac{\ln{\left(7 \right)}}{x}} + \ln{\left(7 \right)} \operatorname{Ei}{\left({\color{red}{u}} \right)} = x e^{- \frac{\ln{\left(7 \right)}}{x}} + \ln{\left(7 \right)} \operatorname{Ei}{\left({\color{red}{\left(- \frac{\ln{\left(7 \right)}}{x}\right)}} \right)}$$

Portanto,

$$\int{7^{- \frac{1}{x}} d x} = x e^{- \frac{\ln{\left(7 \right)}}{x}} + \ln{\left(7 \right)} \operatorname{Ei}{\left(- \frac{\ln{\left(7 \right)}}{x} \right)}$$

Adicione a constante de integração:

$$\int{7^{- \frac{1}{x}} d x} = x e^{- \frac{\ln{\left(7 \right)}}{x}} + \ln{\left(7 \right)} \operatorname{Ei}{\left(- \frac{\ln{\left(7 \right)}}{x} \right)}+C$$

$$$\int 7^{- \frac{1}{x}}\, dx = \left(x e^{- \frac{\ln\left(7\right)}{x}} + \ln\left(7\right) \operatorname{Ei}{\left(- \frac{\ln\left(7\right)}{x} \right)}\right) + C$$$A