$$$\frac{1}{- c + z}$$$ 关于$$$c$$$的积分

求$$$\int \frac{1}{- c + z}\, dc$$$。

设$$$u=- c + z$$$。

则$$$du=\left(- c + z\right)^{\prime }dc = - dc$$$ (步骤见»)，并有$$$dc = - du$$$。

所以，

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

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

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

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

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

回忆一下 $$$u=- c + z$$$:

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - \ln{\left(\left|{{\color{red}{\left(- c + z\right)}}}\right| \right)}$$

因此，

$$\int{\frac{1}{- c + z} d c} = - \ln{\left(\left|{c - z}\right| \right)}$$

加上积分常数：

$$\int{\frac{1}{- c + z} d c} = - \ln{\left(\left|{c - z}\right| \right)}+C$$

$$$\int \frac{1}{- c + z}\, dc = - \ln\left(\left|{c - z}\right|\right) + C$$$A