$$$\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}}$$$ 的积分

求$$$\int \frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}}\, dx$$$。

由于分子次数不小于分母次数，进行多项式长除法（步骤见»）:

$${\color{red}{\int{\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} d x}}} = {\color{red}{\int{\left(\frac{x}{\left(x - 1\right)^{2}} - 2\right)d x}}}$$

逐项积分：

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

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=2$$$：

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

将被积函数的分子改写为 $$$x=x - 1+1$$$，并将分式拆分:

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

逐项积分：

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

设$$$u=x - 1$$$。

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

该积分可以改写为

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

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

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

回忆一下 $$$u=x - 1$$$:

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

设$$$u=x - 1$$$。

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

因此，

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

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

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

回忆一下 $$$u=x - 1$$$:

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

因此，

$$\int{\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} d x} = - 2 x + \ln{\left(\left|{x - 1}\right| \right)} - \frac{1}{x - 1}$$

化简：

$$\int{\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} d x} = \frac{\left(- 2 x + \ln{\left(\left|{x - 1}\right| \right)}\right) \left(x - 1\right) - 1}{x - 1}$$

加上积分常数：

$$\int{\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} d x} = \frac{\left(- 2 x + \ln{\left(\left|{x - 1}\right| \right)}\right) \left(x - 1\right) - 1}{x - 1}+C$$

$$$\int \frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}}\, dx = \frac{\left(- 2 x + \ln\left(\left|{x - 1}\right|\right)\right) \left(x - 1\right) - 1}{x - 1} + C$$$A