Integral de $$$\ln\left(\frac{2}{x}\right)$$$

Encontre $$$\int \ln\left(\frac{2}{x}\right)\, dx$$$.

Para a integral $$$\int{\ln{\left(\frac{2}{x} \right)} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=\ln{\left(\frac{2}{x} \right)}$$$ e $$$\operatorname{dv}=dx$$$.

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

Assim,

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

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=-1$$$:

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

Portanto,

$$\int{\ln{\left(\frac{2}{x} \right)} d x} = x \ln{\left(\frac{2}{x} \right)} + x$$

Simplifique:

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

Adicione a constante de integração:

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

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