$$$\frac{1}{x^{2} - x}$$$ 的积分

求$$$\int \frac{1}{x^{2} - x}\, dx$$$。

进行部分分式分解（步骤可见»）:

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

逐项积分：

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

设$$$u=x - 1$$$。

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

所以，

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

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

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

回忆一下 $$$u=x - 1$$$:

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

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

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

因此，

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

加上积分常数：

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

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