$$$- \frac{1}{u}$$$ 的積分

求$$$\int \left(- \frac{1}{u}\right)\, du$$$。

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=-1$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

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

因此，

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

加上積分常數：

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

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