$$$\frac{2 - x}{1 - x}$$$ 的积分

求$$$\int \frac{2 - x}{1 - x}\, dx$$$。

设$$$u=1 - x$$$。

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

因此，

$${\color{red}{\int{\frac{2 - x}{1 - x} d x}}} = {\color{red}{\int{\left(- \frac{u + 1}{u}\right)d u}}}$$

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

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

Expand the expression:

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

逐项积分：

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

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

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

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

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

回忆一下 $$$u=1 - x$$$:

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

因此，

$$\int{\frac{2 - x}{1 - x} d x} = x - \ln{\left(\left|{x - 1}\right| \right)} - 1$$

加上积分常数（并从表达式中去除常数项）：

$$\int{\frac{2 - x}{1 - x} d x} = x - \ln{\left(\left|{x - 1}\right| \right)}+C$$

$$$\int \frac{2 - x}{1 - x}\, dx = \left(x - \ln\left(\left|{x - 1}\right|\right)\right) + C$$$A