$$$-2 + \frac{1}{u^{2}}$$$の積分

$$$\int \left(-2 + \frac{1}{u^{2}}\right)\, du$$$ を求めよ。

項別に積分せよ:

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

$$$c=2$$$ に対して定数則 $$$\int c\, du = c u$$$ を適用する:

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

$$$n=-2$$$ を用いて、べき乗の法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

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

したがって、

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

積分定数を加える:

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

$$$\int \left(-2 + \frac{1}{u^{2}}\right)\, du = \left(- 2 u - \frac{1}{u}\right) + C$$$A