$$$- \frac{1}{t}$$$の積分

$$$\int \left(- \frac{1}{t}\right)\, dt$$$ 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、$$$c=-1$$$ と $$$f{\left(t \right)} = \frac{1}{t}$$$ に対して適用する:

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

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

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

したがって、

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

積分定数を加える:

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

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