$$$\frac{1}{a^{2} u}$$$ 對 $$$u$$$ 的積分

求$$$\int \frac{1}{a^{2} u}\, du$$$。

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{a^{2}}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

$${\color{red}{\int{\frac{1}{a^{2} u} d u}}} = {\color{red}{\frac{\int{\frac{1}{u} d u}}{a^{2}}}}$$

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

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

因此，

$$\int{\frac{1}{a^{2} u} d u} = \frac{\ln{\left(\left|{u}\right| \right)}}{a^{2}}$$

加上積分常數：

$$\int{\frac{1}{a^{2} u} d u} = \frac{\ln{\left(\left|{u}\right| \right)}}{a^{2}}+C$$

$$$\int \frac{1}{a^{2} u}\, du = \frac{\ln\left(\left|{u}\right|\right)}{a^{2}} + C$$$A