$$$\frac{6}{r}$$$の積分

$$$\int \frac{6}{r}\, dr$$$ を求めよ。

定数倍の法則 $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$ を、$$$c=6$$$ と $$$f{\left(r \right)} = \frac{1}{r}$$$ に対して適用する:

$${\color{red}{\int{\frac{6}{r} d r}}} = {\color{red}{\left(6 \int{\frac{1}{r} d r}\right)}}$$

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

$$6 {\color{red}{\int{\frac{1}{r} d r}}} = 6 {\color{red}{\ln{\left(\left|{r}\right| \right)}}}$$

したがって、

$$\int{\frac{6}{r} d r} = 6 \ln{\left(\left|{r}\right| \right)}$$

積分定数を加える:

$$\int{\frac{6}{r} d r} = 6 \ln{\left(\left|{r}\right| \right)}+C$$

$$$\int \frac{6}{r}\, dr = 6 \ln\left(\left|{r}\right|\right) + C$$$A