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

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

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

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

Seja $$$u=- x$$$.

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

Logo,

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

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

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

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

Sejam $$$\operatorname{g}=\ln{\left(u \right)}$$$ e $$$\operatorname{dv}=du$$$.

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

A integral torna-se

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

Aplique a regra da constante $$$\int c\, du = c u$$$ usando $$$c=1$$$:

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

Recorde que $$$u=- x$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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