$$$\frac{1}{\left(x^{2} - 20 x\right)^{2}}$$$ 的积分

求$$$\int \frac{1}{\left(x^{2} - 20 x\right)^{2}}\, dx$$$。

进行部分分式分解（步骤可见»）:

$${\color{red}{\int{\frac{1}{\left(x^{2} - 20 x\right)^{2}} d x}}} = {\color{red}{\int{\left(- \frac{1}{4000 \left(x - 20\right)} + \frac{1}{400 \left(x - 20\right)^{2}} + \frac{1}{4000 x} + \frac{1}{400 x^{2}}\right)d x}}}$$

逐项积分：

$${\color{red}{\int{\left(- \frac{1}{4000 \left(x - 20\right)} + \frac{1}{400 \left(x - 20\right)^{2}} + \frac{1}{4000 x} + \frac{1}{400 x^{2}}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{400 x^{2}} d x} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} - \int{\frac{1}{4000 \left(x - 20\right)} d x}\right)}}$$

对 $$$c=\frac{1}{4000}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x - 20}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\int{\frac{1}{400 x^{2}} d x} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} - {\color{red}{\int{\frac{1}{4000 \left(x - 20\right)} d x}}} = \int{\frac{1}{400 x^{2}} d x} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x - 20} d x}}{4000}\right)}}$$

设$$$u=x - 20$$$。

则$$$du=\left(x - 20\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

因此，

$$\int{\frac{1}{400 x^{2}} d x} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} - \frac{{\color{red}{\int{\frac{1}{x - 20} d x}}}}{4000} = \int{\frac{1}{400 x^{2}} d x} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4000}$$

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

$$\int{\frac{1}{400 x^{2}} d x} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4000} = \int{\frac{1}{400 x^{2}} d x} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4000}$$

回忆一下 $$$u=x - 20$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4000} + \int{\frac{1}{400 x^{2}} d x} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x - 20\right)}}}\right| \right)}}{4000} + \int{\frac{1}{400 x^{2}} d x} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x}$$

对 $$$c=\frac{1}{400}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} + {\color{red}{\int{\frac{1}{400 x^{2}} d x}}} = - \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} + {\color{red}{\left(\frac{\int{\frac{1}{x^{2}} d x}}{400}\right)}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-2$$$：

$$- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} + \frac{{\color{red}{\int{\frac{1}{x^{2}} d x}}}}{400}=- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} + \frac{{\color{red}{\int{x^{-2} d x}}}}{400}=- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} + \frac{{\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}}{400}=- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} + \frac{{\color{red}{\left(- x^{-1}\right)}}}{400}=- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \int{\frac{1}{400 \left(x - 20\right)^{2}} d x} + \frac{{\color{red}{\left(- \frac{1}{x}\right)}}}{400}$$

对 $$$c=\frac{1}{400}$$$ 和 $$$f{\left(x \right)} = \frac{1}{\left(x - 20\right)^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + {\color{red}{\int{\frac{1}{400 \left(x - 20\right)^{2}} d x}}} - \frac{1}{400 x} = - \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + {\color{red}{\left(\frac{\int{\frac{1}{\left(x - 20\right)^{2}} d x}}{400}\right)}} - \frac{1}{400 x}$$

设$$$u=x - 20$$$。

则$$$du=\left(x - 20\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

所以，

$$- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \frac{{\color{red}{\int{\frac{1}{\left(x - 20\right)^{2}} d x}}}}{400} - \frac{1}{400 x} = - \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{400} - \frac{1}{400 x}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-2$$$：

$$- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{400} - \frac{1}{400 x}=- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \frac{{\color{red}{\int{u^{-2} d u}}}}{400} - \frac{1}{400 x}=- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \frac{{\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{400} - \frac{1}{400 x}=- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \frac{{\color{red}{\left(- u^{-1}\right)}}}{400} - \frac{1}{400 x}=- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} + \frac{{\color{red}{\left(- \frac{1}{u}\right)}}}{400} - \frac{1}{400 x}$$

回忆一下 $$$u=x - 20$$$:

$$- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} - \frac{{\color{red}{u}}^{-1}}{400} - \frac{1}{400 x} = - \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \int{\frac{1}{4000 x} d x} - \frac{{\color{red}{\left(x - 20\right)}}^{-1}}{400} - \frac{1}{400 x}$$

对 $$$c=\frac{1}{4000}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + {\color{red}{\int{\frac{1}{4000 x} d x}}} - \frac{1}{400 \left(x - 20\right)} - \frac{1}{400 x} = - \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + {\color{red}{\left(\frac{\int{\frac{1}{x} d x}}{4000}\right)}} - \frac{1}{400 \left(x - 20\right)} - \frac{1}{400 x}$$

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

$$- \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \frac{{\color{red}{\int{\frac{1}{x} d x}}}}{4000} - \frac{1}{400 \left(x - 20\right)} - \frac{1}{400 x} = - \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} + \frac{{\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{4000} - \frac{1}{400 \left(x - 20\right)} - \frac{1}{400 x}$$

因此，

$$\int{\frac{1}{\left(x^{2} - 20 x\right)^{2}} d x} = \frac{\ln{\left(\left|{x}\right| \right)}}{4000} - \frac{\ln{\left(\left|{x - 20}\right| \right)}}{4000} - \frac{1}{400 \left(x - 20\right)} - \frac{1}{400 x}$$

化简：

$$\int{\frac{1}{\left(x^{2} - 20 x\right)^{2}} d x} = \frac{x \left(x - 20\right) \left(\ln{\left(\left|{x}\right| \right)} - \ln{\left(\left|{x - 20}\right| \right)}\right) - 20 x + 200}{4000 x \left(x - 20\right)}$$

加上积分常数：

$$\int{\frac{1}{\left(x^{2} - 20 x\right)^{2}} d x} = \frac{x \left(x - 20\right) \left(\ln{\left(\left|{x}\right| \right)} - \ln{\left(\left|{x - 20}\right| \right)}\right) - 20 x + 200}{4000 x \left(x - 20\right)}+C$$

$$$\int \frac{1}{\left(x^{2} - 20 x\right)^{2}}\, dx = \frac{x \left(x - 20\right) \left(\ln\left(\left|{x}\right|\right) - \ln\left(\left|{x - 20}\right|\right)\right) - 20 x + 200}{4000 x \left(x - 20\right)} + C$$$A