$$$\frac{1}{x^{2} \left(x - 1\right)}$$$ 的積分

求$$$\int \frac{1}{x^{2} \left(x - 1\right)}\, dx$$$。

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

令 $$$u=x - 1$$$。

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

因此，

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

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

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

回顧一下 $$$u=x - 1$$$：

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

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

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

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

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

因此，

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

加上積分常數：

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

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