$$$\frac{1}{a^{2} u}$$$ の $$$u$$$ に関する積分
入力内容
$$$\int \frac{1}{a^{2} u}\, du$$$ を求めよ。
解答
定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{a^{2}}$$$ と $$$f{\left(u \right)} = \frac{1}{u}$$$ に対して適用する:
$${\color{red}{\int{\frac{1}{a^{2} u} d u}}} = {\color{red}{\frac{\int{\frac{1}{u} d u}}{a^{2}}}}$$
$$$\frac{1}{u}$$$ の不定積分は $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$ です:
$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{a^{2}} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{a^{2}}$$
したがって、
$$\int{\frac{1}{a^{2} u} d u} = \frac{\ln{\left(\left|{u}\right| \right)}}{a^{2}}$$
積分定数を加える:
$$\int{\frac{1}{a^{2} u} d u} = \frac{\ln{\left(\left|{u}\right| \right)}}{a^{2}}+C$$
解答
$$$\int \frac{1}{a^{2} u}\, du = \frac{\ln\left(\left|{u}\right|\right)}{a^{2}} + C$$$A