$$$\frac{4 x}{\left(x - 2\right)^{2}}$$$ 的积分

求$$$\int \frac{4 x}{\left(x - 2\right)^{2}}\, dx$$$。

对 $$$c=4$$$ 和 $$$f{\left(x \right)} = \frac{x}{\left(x - 2\right)^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{4 x}{\left(x - 2\right)^{2}} d x}}} = {\color{red}{\left(4 \int{\frac{x}{\left(x - 2\right)^{2}} d x}\right)}}$$

将被积函数的分子改写为 $$$x=x - 2+2$$$，并将分式拆分:

$$4 {\color{red}{\int{\frac{x}{\left(x - 2\right)^{2}} d x}}} = 4 {\color{red}{\int{\left(\frac{1}{x - 2} + \frac{2}{\left(x - 2\right)^{2}}\right)d x}}}$$

逐项积分：

$$4 {\color{red}{\int{\left(\frac{1}{x - 2} + \frac{2}{\left(x - 2\right)^{2}}\right)d x}}} = 4 {\color{red}{\left(\int{\frac{2}{\left(x - 2\right)^{2}} d x} + \int{\frac{1}{x - 2} d x}\right)}}$$

设$$$u=x - 2$$$。

则$$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

因此，

$$4 \int{\frac{2}{\left(x - 2\right)^{2}} d x} + 4 {\color{red}{\int{\frac{1}{x - 2} d x}}} = 4 \int{\frac{2}{\left(x - 2\right)^{2}} d x} + 4 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$4 \int{\frac{2}{\left(x - 2\right)^{2}} d x} + 4 {\color{red}{\int{\frac{1}{u} d u}}} = 4 \int{\frac{2}{\left(x - 2\right)^{2}} d x} + 4 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回忆一下 $$$u=x - 2$$$:

$$4 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + 4 \int{\frac{2}{\left(x - 2\right)^{2}} d x} = 4 \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)} + 4 \int{\frac{2}{\left(x - 2\right)^{2}} d x}$$

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

$$4 \ln{\left(\left|{x - 2}\right| \right)} + 4 {\color{red}{\int{\frac{2}{\left(x - 2\right)^{2}} d x}}} = 4 \ln{\left(\left|{x - 2}\right| \right)} + 4 {\color{red}{\left(2 \int{\frac{1}{\left(x - 2\right)^{2}} d x}\right)}}$$

设$$$u=x - 2$$$。

则$$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

该积分可以改写为

$$4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\int{\frac{1}{\left(x - 2\right)^{2}} d x}}} = 4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\int{\frac{1}{u^{2}} d u}}}$$

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

$$4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\int{u^{-2} d u}}}=4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\left(- u^{-1}\right)}}=4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\left(- \frac{1}{u}\right)}}$$

回忆一下 $$$u=x - 2$$$:

$$4 \ln{\left(\left|{x - 2}\right| \right)} - 8 {\color{red}{u}}^{-1} = 4 \ln{\left(\left|{x - 2}\right| \right)} - 8 {\color{red}{\left(x - 2\right)}}^{-1}$$

因此，

$$\int{\frac{4 x}{\left(x - 2\right)^{2}} d x} = 4 \ln{\left(\left|{x - 2}\right| \right)} - \frac{8}{x - 2}$$

化简：

$$\int{\frac{4 x}{\left(x - 2\right)^{2}} d x} = \frac{4 \left(\left(x - 2\right) \ln{\left(\left|{x - 2}\right| \right)} - 2\right)}{x - 2}$$

加上积分常数：

$$\int{\frac{4 x}{\left(x - 2\right)^{2}} d x} = \frac{4 \left(\left(x - 2\right) \ln{\left(\left|{x - 2}\right| \right)} - 2\right)}{x - 2}+C$$

$$$\int \frac{4 x}{\left(x - 2\right)^{2}}\, dx = \frac{4 \left(\left(x - 2\right) \ln\left(\left|{x - 2}\right|\right) - 2\right)}{x - 2} + C$$$A