$$$\frac{1}{u^{2} - 2 u}$$$ 的积分

求$$$\int \frac{1}{u^{2} - 2 u}\, du$$$。

进行部分分式分解（步骤可见»）:

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

逐项积分：

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

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

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

设$$$v=u - 2$$$。

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

该积分可以改写为

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

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

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

回忆一下 $$$v=u - 2$$$:

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

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

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

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

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

因此，

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

化简：

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

加上积分常数：

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

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