$$$\frac{2}{x - 2}$$$ 的积分

求$$$\int \frac{2}{x - 2}\, dx$$$。

对 $$$c=2$$$ 和 $$$f{\left(x \right)} = \frac{1}{x - 2}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

设$$$u=x - 2$$$。

则$$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

该积分可以改写为

$$2 {\color{red}{\int{\frac{1}{x - 2} d x}}} = 2 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$2 {\color{red}{\int{\frac{1}{u} d u}}} = 2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回忆一下 $$$u=x - 2$$$:

$$2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 2 \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)}$$

因此，

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

加上积分常数：

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

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