$$$\frac{1}{f \left(2 a - x\right)}$$$ 关于$$$x$$$的积分

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

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

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

设$$$u=2 a - x$$$。

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

因此，

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

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

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

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

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

回忆一下 $$$u=2 a - x$$$:

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

因此，

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

加上积分常数：

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

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