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

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

Alterar a base:

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

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

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

Assim,

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

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

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

Seja $$$v=u \ln{\left(2 \right)}$$$.

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

A integral torna-se

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

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

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

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

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

Recorde que $$$v=u \ln{\left(2 \right)}$$$:

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

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

$$- \frac{e^{\ln{\left(2 \right)} {\color{red}{u}}}}{\ln{\left(2 \right)}} = - \frac{e^{\ln{\left(2 \right)} {\color{red}{\frac{1}{x}}}}}{\ln{\left(2 \right)}}$$

Portanto,

$$\int{\frac{2^{\frac{1}{x}}}{x^{2}} d x} = - \frac{e^{\frac{\ln{\left(2 \right)}}{x}}}{\ln{\left(2 \right)}}$$

Adicione a constante de integração:

$$\int{\frac{2^{\frac{1}{x}}}{x^{2}} d x} = - \frac{e^{\frac{\ln{\left(2 \right)}}{x}}}{\ln{\left(2 \right)}}+C$$

$$$\int \frac{2^{\frac{1}{x}}}{x^{2}}\, dx = - \frac{e^{\frac{\ln\left(2\right)}{x}}}{\ln\left(2\right)} + C$$$A