$$$x^{2} \left(x - 5\right)^{13}$$$の積分

$$$\int x^{2} \left(x - 5\right)^{13}\, dx$$$ を求めよ。

$$$u=x - 5$$$ とする。

すると $$$du=\left(x - 5\right)^{\prime }dx = 1 dx$$$（手順は»で確認できます）、$$$dx = du$$$ となります。

したがって、

$${\color{red}{\int{x^{2} \left(x - 5\right)^{13} d x}}} = {\color{red}{\int{u^{13} \left(u + 5\right)^{2} d u}}}$$

Expand the expression:

$${\color{red}{\int{u^{13} \left(u + 5\right)^{2} d u}}} = {\color{red}{\int{\left(u^{15} + 10 u^{14} + 25 u^{13}\right)d u}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(u^{15} + 10 u^{14} + 25 u^{13}\right)d u}}} = {\color{red}{\left(\int{25 u^{13} d u} + \int{10 u^{14} d u} + \int{u^{15} d u}\right)}}$$

$$$n=15$$$ 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:

$$\int{25 u^{13} d u} + \int{10 u^{14} d u} + {\color{red}{\int{u^{15} d u}}}=\int{25 u^{13} d u} + \int{10 u^{14} d u} + {\color{red}{\frac{u^{1 + 15}}{1 + 15}}}=\int{25 u^{13} d u} + \int{10 u^{14} d u} + {\color{red}{\left(\frac{u^{16}}{16}\right)}}$$

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=10$$$ と $$$f{\left(u \right)} = u^{14}$$$ に対して適用する:

$$\frac{u^{16}}{16} + \int{25 u^{13} d u} + {\color{red}{\int{10 u^{14} d u}}} = \frac{u^{16}}{16} + \int{25 u^{13} d u} + {\color{red}{\left(10 \int{u^{14} d u}\right)}}$$

$$$n=14$$$ 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:

$$\frac{u^{16}}{16} + \int{25 u^{13} d u} + 10 {\color{red}{\int{u^{14} d u}}}=\frac{u^{16}}{16} + \int{25 u^{13} d u} + 10 {\color{red}{\frac{u^{1 + 14}}{1 + 14}}}=\frac{u^{16}}{16} + \int{25 u^{13} d u} + 10 {\color{red}{\left(\frac{u^{15}}{15}\right)}}$$

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=25$$$ と $$$f{\left(u \right)} = u^{13}$$$ に対して適用する:

$$\frac{u^{16}}{16} + \frac{2 u^{15}}{3} + {\color{red}{\int{25 u^{13} d u}}} = \frac{u^{16}}{16} + \frac{2 u^{15}}{3} + {\color{red}{\left(25 \int{u^{13} d u}\right)}}$$

$$$n=13$$$ 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:

$$\frac{u^{16}}{16} + \frac{2 u^{15}}{3} + 25 {\color{red}{\int{u^{13} d u}}}=\frac{u^{16}}{16} + \frac{2 u^{15}}{3} + 25 {\color{red}{\frac{u^{1 + 13}}{1 + 13}}}=\frac{u^{16}}{16} + \frac{2 u^{15}}{3} + 25 {\color{red}{\left(\frac{u^{14}}{14}\right)}}$$

次のことを思い出してください $$$u=x - 5$$$:

$$\frac{25 {\color{red}{u}}^{14}}{14} + \frac{2 {\color{red}{u}}^{15}}{3} + \frac{{\color{red}{u}}^{16}}{16} = \frac{25 {\color{red}{\left(x - 5\right)}}^{14}}{14} + \frac{2 {\color{red}{\left(x - 5\right)}}^{15}}{3} + \frac{{\color{red}{\left(x - 5\right)}}^{16}}{16}$$

したがって、

$$\int{x^{2} \left(x - 5\right)^{13} d x} = \frac{\left(x - 5\right)^{16}}{16} + \frac{2 \left(x - 5\right)^{15}}{3} + \frac{25 \left(x - 5\right)^{14}}{14}$$

簡単化せよ:

$$\int{x^{2} \left(x - 5\right)^{13} d x} = \frac{\left(x - 5\right)^{14} \left(224 x + 21 \left(x - 5\right)^{2} - 520\right)}{336}$$

積分定数を加える:

$$\int{x^{2} \left(x - 5\right)^{13} d x} = \frac{\left(x - 5\right)^{14} \left(224 x + 21 \left(x - 5\right)^{2} - 520\right)}{336}+C$$

$$$\int x^{2} \left(x - 5\right)^{13}\, dx = \frac{\left(x - 5\right)^{14} \left(224 x + 21 \left(x - 5\right)^{2} - 520\right)}{336} + C$$$A