$$$\left(1 - u^{2}\right)^{2}$$$の積分

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

Expand the expression:

$${\color{red}{\int{\left(1 - u^{2}\right)^{2} d u}}} = {\color{red}{\int{\left(u^{4} - 2 u^{2} + 1\right)d u}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(u^{4} - 2 u^{2} + 1\right)d u}}} = {\color{red}{\left(\int{1 d u} - \int{2 u^{2} d u} + \int{u^{4} d u}\right)}}$$

$$$c=1$$$ に対して定数則 $$$\int c\, du = c u$$$ を適用する:

$$- \int{2 u^{2} d u} + \int{u^{4} d u} + {\color{red}{\int{1 d u}}} = - \int{2 u^{2} d u} + \int{u^{4} d u} + {\color{red}{u}}$$

$$$n=4$$$ 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:

$$u - \int{2 u^{2} d u} + {\color{red}{\int{u^{4} d u}}}=u - \int{2 u^{2} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=u - \int{2 u^{2} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=2$$$ と $$$f{\left(u \right)} = u^{2}$$$ に対して適用する:

$$\frac{u^{5}}{5} + u - {\color{red}{\int{2 u^{2} d u}}} = \frac{u^{5}}{5} + u - {\color{red}{\left(2 \int{u^{2} d u}\right)}}$$

$$$n=2$$$ 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:

$$\frac{u^{5}}{5} + u - 2 {\color{red}{\int{u^{2} d u}}}=\frac{u^{5}}{5} + u - 2 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=\frac{u^{5}}{5} + u - 2 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

したがって、

$$\int{\left(1 - u^{2}\right)^{2} d u} = \frac{u^{5}}{5} - \frac{2 u^{3}}{3} + u$$

積分定数を加える:

$$\int{\left(1 - u^{2}\right)^{2} d u} = \frac{u^{5}}{5} - \frac{2 u^{3}}{3} + u+C$$

$$$\int \left(1 - u^{2}\right)^{2}\, du = \left(\frac{u^{5}}{5} - \frac{2 u^{3}}{3} + u\right) + C$$$A