$$$- 3 x_{2} + \frac{1}{x}$$$ の $$$x$$$ に関する積分

$$$\int \left(- 3 x_{2} + \frac{1}{x}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(- 3 x_{2} + \frac{1}{x}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{x} d x} - \int{3 x_{2} d x}\right)}}$$

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

$$- \int{3 x_{2} d x} + {\color{red}{\int{\frac{1}{x} d x}}} = - \int{3 x_{2} d x} + {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

$$$c=3 x_{2}$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$\ln{\left(\left|{x}\right| \right)} - {\color{red}{\int{3 x_{2} d x}}} = \ln{\left(\left|{x}\right| \right)} - {\color{red}{\left(3 x x_{2}\right)}}$$

したがって、

$$\int{\left(- 3 x_{2} + \frac{1}{x}\right)d x} = - 3 x x_{2} + \ln{\left(\left|{x}\right| \right)}$$

積分定数を加える:

$$\int{\left(- 3 x_{2} + \frac{1}{x}\right)d x} = - 3 x x_{2} + \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \left(- 3 x_{2} + \frac{1}{x}\right)\, dx = \left(- 3 x x_{2} + \ln\left(\left|{x}\right|\right)\right) + C$$$A