$$$\ln\left(-1 + \frac{1}{x}\right)$$$ 的积分

求$$$\int \ln\left(-1 + \frac{1}{x}\right)\, dx$$$。

对于积分$$$\int{\ln{\left(-1 + \frac{1}{x} \right)} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=\ln{\left(-1 + \frac{1}{x} \right)}$$$ 和 $$$\operatorname{dv}=dx$$$。

则 $$$\operatorname{du}=\left(\ln{\left(-1 + \frac{1}{x} \right)}\right)^{\prime }dx=\frac{1}{x \left(x - 1\right)} dx$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d x}=x$$$ (步骤见 »)。

积分变为

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

设$$$u=x - 1$$$。

则$$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

所以，

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

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

回忆一下 $$$u=x - 1$$$:

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

因此，

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

化简：

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

加上积分常数：

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

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