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

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=2$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{2} \left(1 - x\right)}$$$：

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

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

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

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

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

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

令 $$$u=x - 1$$$。

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

因此，

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

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

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

回顧一下 $$$u=x - 1$$$：

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

因此，

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

化簡：

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

加上積分常數：

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

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