$$$x - \frac{1}{x}$$$の積分

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

項別に積分せよ:

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

$$$n=1$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

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

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

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

したがって、

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

積分定数を加える:

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

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