$$$- \frac{x}{x - 1}$$$ 的积分

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

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

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

改写并拆分该分式:

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

逐项积分：

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

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

$$- \int{\frac{1}{x - 1} d x} - {\color{red}{\int{1 d x}}} = - \int{\frac{1}{x - 1} d x} - {\color{red}{x}}$$

设$$$u=x - 1$$$。

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

所以，

$$- x - {\color{red}{\int{\frac{1}{x - 1} d x}}} = - x - {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

回忆一下 $$$u=x - 1$$$:

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

因此，

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

加上积分常数：

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

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