$$$- u^{2} + \frac{1}{a^{2}}$$$ の $$$u$$$ に関する積分

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

項別に積分せよ:

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

$$$c=\frac{1}{a^{2}}$$$ に対して定数則 $$$\int c\, du = c u$$$ を適用する:

$$- \int{u^{2} d u} + {\color{red}{\int{\frac{1}{a^{2}} d u}}} = - \int{u^{2} d u} + {\color{red}{\frac{u}{a^{2}}}}$$

$$$n=2$$$ 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:

$$- {\color{red}{\int{u^{2} d u}}} + \frac{u}{a^{2}}=- {\color{red}{\frac{u^{1 + 2}}{1 + 2}}} + \frac{u}{a^{2}}=- {\color{red}{\left(\frac{u^{3}}{3}\right)}} + \frac{u}{a^{2}}$$

したがって、

$$\int{\left(- u^{2} + \frac{1}{a^{2}}\right)d u} = - \frac{u^{3}}{3} + \frac{u}{a^{2}}$$

積分定数を加える:

$$\int{\left(- u^{2} + \frac{1}{a^{2}}\right)d u} = - \frac{u^{3}}{3} + \frac{u}{a^{2}}+C$$

$$$\int \left(- u^{2} + \frac{1}{a^{2}}\right)\, du = \left(- \frac{u^{3}}{3} + \frac{u}{a^{2}}\right) + C$$$A