$$$\left(9 - x^{2}\right)^{2} - \left(x + 7\right)^{2}$$$の積分

$$$\int \left(\left(9 - x^{2}\right)^{2} - \left(x + 7\right)^{2}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(\left(9 - x^{2}\right)^{2} - \left(x + 7\right)^{2}\right)d x}}} = {\color{red}{\left(\int{\left(9 - x^{2}\right)^{2} d x} - \int{\left(x + 7\right)^{2} d x}\right)}}$$

Expand the expression:

$$- \int{\left(x + 7\right)^{2} d x} + {\color{red}{\int{\left(9 - x^{2}\right)^{2} d x}}} = - \int{\left(x + 7\right)^{2} d x} + {\color{red}{\int{\left(x^{4} - 18 x^{2} + 81\right)d x}}}$$

項別に積分せよ:

$$- \int{\left(x + 7\right)^{2} d x} + {\color{red}{\int{\left(x^{4} - 18 x^{2} + 81\right)d x}}} = - \int{\left(x + 7\right)^{2} d x} + {\color{red}{\left(\int{81 d x} - \int{18 x^{2} d x} + \int{x^{4} d x}\right)}}$$

$$$c=81$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$- \int{18 x^{2} d x} + \int{x^{4} d x} - \int{\left(x + 7\right)^{2} d x} + {\color{red}{\int{81 d x}}} = - \int{18 x^{2} d x} + \int{x^{4} d x} - \int{\left(x + 7\right)^{2} d x} + {\color{red}{\left(81 x\right)}}$$

$$$n=4$$$ 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:

$$81 x - \int{18 x^{2} d x} - \int{\left(x + 7\right)^{2} d x} + {\color{red}{\int{x^{4} d x}}}=81 x - \int{18 x^{2} d x} - \int{\left(x + 7\right)^{2} d x} + {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=81 x - \int{18 x^{2} d x} - \int{\left(x + 7\right)^{2} d x} + {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=18$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用する:

$$\frac{x^{5}}{5} + 81 x - \int{\left(x + 7\right)^{2} d x} - {\color{red}{\int{18 x^{2} d x}}} = \frac{x^{5}}{5} + 81 x - \int{\left(x + 7\right)^{2} d x} - {\color{red}{\left(18 \int{x^{2} d x}\right)}}$$

$$$n=2$$$ 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:

$$\frac{x^{5}}{5} + 81 x - \int{\left(x + 7\right)^{2} d x} - 18 {\color{red}{\int{x^{2} d x}}}=\frac{x^{5}}{5} + 81 x - \int{\left(x + 7\right)^{2} d x} - 18 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\frac{x^{5}}{5} + 81 x - \int{\left(x + 7\right)^{2} d x} - 18 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

$$$u=x + 7$$$ とする。

すると $$$du=\left(x + 7\right)^{\prime }dx = 1 dx$$$（手順は»で確認できます）、$$$dx = du$$$ となります。

したがって、

$$\frac{x^{5}}{5} - 6 x^{3} + 81 x - {\color{red}{\int{\left(x + 7\right)^{2} d x}}} = \frac{x^{5}}{5} - 6 x^{3} + 81 x - {\color{red}{\int{u^{2} d u}}}$$

$$$n=2$$$ 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:

$$\frac{x^{5}}{5} - 6 x^{3} + 81 x - {\color{red}{\int{u^{2} d u}}}=\frac{x^{5}}{5} - 6 x^{3} + 81 x - {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=\frac{x^{5}}{5} - 6 x^{3} + 81 x - {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

次のことを思い出してください $$$u=x + 7$$$:

$$\frac{x^{5}}{5} - 6 x^{3} + 81 x - \frac{{\color{red}{u}}^{3}}{3} = \frac{x^{5}}{5} - 6 x^{3} + 81 x - \frac{{\color{red}{\left(x + 7\right)}}^{3}}{3}$$

したがって、

$$\int{\left(\left(9 - x^{2}\right)^{2} - \left(x + 7\right)^{2}\right)d x} = \frac{x^{5}}{5} - 6 x^{3} + 81 x - \frac{\left(x + 7\right)^{3}}{3}$$

積分定数を加える:

$$\int{\left(\left(9 - x^{2}\right)^{2} - \left(x + 7\right)^{2}\right)d x} = \frac{x^{5}}{5} - 6 x^{3} + 81 x - \frac{\left(x + 7\right)^{3}}{3}+C$$

$$$\int \left(\left(9 - x^{2}\right)^{2} - \left(x + 7\right)^{2}\right)\, dx = \left(\frac{x^{5}}{5} - 6 x^{3} + 81 x - \frac{\left(x + 7\right)^{3}}{3}\right) + C$$$A