$$$\frac{1}{- a + t}$$$ 关于$$$t$$$的积分

求$$$\int \frac{1}{- a + t}\, dt$$$。

设$$$u=- a + t$$$。

则$$$du=\left(- a + t\right)^{\prime }dt = 1 dt$$$ (步骤见»)，并有$$$dt = du$$$。

所以，

$${\color{red}{\int{\frac{1}{- a + t} d t}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

回忆一下 $$$u=- a + t$$$:

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

因此，

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

加上积分常数：

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

$$$\int \frac{1}{- a + t}\, dt = \ln\left(\left|{a - t}\right|\right) + C$$$A