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

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

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

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

Assim,

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

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

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

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

Logo,

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

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 $$$u=x - 1$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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