$$$\frac{5}{2 x - 2}$$$ 的积分

求$$$\int \frac{5}{2 x - 2}\, dx$$$。

化简被积函数:

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

对 $$$c=\frac{5}{2}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x - 1}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

设$$$u=x - 1$$$。

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

因此，

$$\frac{5 {\color{red}{\int{\frac{1}{x - 1} d x}}}}{2} = \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$\frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{5 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

回忆一下 $$$u=x - 1$$$:

$$\frac{5 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{5 \ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{2}$$

因此，

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

加上积分常数：

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

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