Integral de $$$\ln\left(1 - x\right)$$$

Encontre $$$\int \ln\left(1 - x\right)\, dx$$$.

Seja $$$u=1 - x$$$.

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

Logo,

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

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

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

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

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

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

Portanto,

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

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

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

Recorde que $$$u=1 - x$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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