Integral de $$$\frac{- 2 \ln\left(x\right) - 4}{x}$$$

Encontre $$$\int \frac{- 2 \ln\left(x\right) - 4}{x}\, dx$$$.

Simplifique o integrando:

$${\color{red}{\int{\frac{- 2 \ln{\left(x \right)} - 4}{x} d x}}} = {\color{red}{\int{\left(- \frac{2 \left(\ln{\left(x \right)} + 2\right)}{x}\right)d x}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=-2$$$ e $$$f{\left(x \right)} = \frac{\ln{\left(x \right)} + 2}{x}$$$:

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

Expand the expression:

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

Integre termo a termo:

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

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

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

A integral de $$$\frac{1}{x}$$$ é $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

$$- 2 \int{\frac{\ln{\left(x \right)}}{x} d x} - 4 {\color{red}{\int{\frac{1}{x} d x}}} = - 2 \int{\frac{\ln{\left(x \right)}}{x} d x} - 4 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

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

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

Portanto,

$$- 4 \ln{\left(\left|{x}\right| \right)} - 2 {\color{red}{\int{\frac{\ln{\left(x \right)}}{x} d x}}} = - 4 \ln{\left(\left|{x}\right| \right)} - 2 {\color{red}{\int{u d u}}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=1$$$:

$$- 4 \ln{\left(\left|{x}\right| \right)} - 2 {\color{red}{\int{u d u}}}=- 4 \ln{\left(\left|{x}\right| \right)} - 2 {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}=- 4 \ln{\left(\left|{x}\right| \right)} - 2 {\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

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

$$- 4 \ln{\left(\left|{x}\right| \right)} - {\color{red}{u}}^{2} = - 4 \ln{\left(\left|{x}\right| \right)} - {\color{red}{\ln{\left(x \right)}}}^{2}$$

Portanto,

$$\int{\frac{- 2 \ln{\left(x \right)} - 4}{x} d x} = - \ln{\left(x \right)}^{2} - 4 \ln{\left(\left|{x}\right| \right)}$$

Adicione a constante de integração:

$$\int{\frac{- 2 \ln{\left(x \right)} - 4}{x} d x} = - \ln{\left(x \right)}^{2} - 4 \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \frac{- 2 \ln\left(x\right) - 4}{x}\, dx = \left(- \ln^{2}\left(x\right) - 4 \ln\left(\left|{x}\right|\right)\right) + C$$$A