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

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

逐项积分：

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

设$$$u=1 - x$$$。

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

因此，

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

对 $$$c=-1$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

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

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

回忆一下 $$$u=1 - x$$$:

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

设$$$u=\ln{\left(x \right)}$$$。

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

积分变为

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

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

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

回忆一下 $$$u=\ln{\left(x \right)}$$$:

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

因此，

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

加上积分常数：

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

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