Integral de $$$\frac{e^{- \frac{1}{x}}}{x^{2}}$$$

Encontre $$$\int \frac{e^{- \frac{1}{x}}}{x^{2}}\, dx$$$.

Seja $$$u=- \frac{1}{x}$$$.

Então $$$du=\left(- \frac{1}{x}\right)^{\prime }dx = \frac{dx}{x^{2}}$$$ (veja os passos »), e obtemos $$$\frac{dx}{x^{2}} = du$$$.

Assim,

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

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

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

Recorde que $$$u=- \frac{1}{x}$$$:

$$e^{{\color{red}{u}}} = e^{{\color{red}{\left(- \frac{1}{x}\right)}}}$$

Portanto,

$$\int{\frac{e^{- \frac{1}{x}}}{x^{2}} d x} = e^{- \frac{1}{x}}$$

Adicione a constante de integração:

$$\int{\frac{e^{- \frac{1}{x}}}{x^{2}} d x} = e^{- \frac{1}{x}}+C$$

$$$\int \frac{e^{- \frac{1}{x}}}{x^{2}}\, dx = e^{- \frac{1}{x}} + C$$$A