$$$\frac{1}{x - 2}$$$ 的積分

求$$$\int \frac{1}{x - 2}\, dx$$$。

令 $$$u=x - 2$$$。

則 $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

該積分可改寫為

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

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

回顧一下 $$$u=x - 2$$$：

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

因此，

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

加上積分常數：

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

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