$$$\frac{e_{1}}{t}$$$ の $$$t$$$ に関する積分

$$$\int \frac{e_{1}}{t}\, dt$$$ を求めよ。

定数倍の法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ を、$$$c=e_{1}$$$ と $$$f{\left(t \right)} = \frac{1}{t}$$$ に対して適用する:

$${\color{red}{\int{\frac{e_{1}}{t} d t}}} = {\color{red}{e_{1} \int{\frac{1}{t} d t}}}$$

$$$\frac{1}{t}$$$ の不定積分は $$$\int{\frac{1}{t} d t} = \ln{\left(\left|{t}\right| \right)}$$$ です:

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

したがって、

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

積分定数を加える:

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

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