$$$x^{2} \left(x - 5\right)^{13}$$$의 적분

$$$\int x^{2} \left(x - 5\right)^{13}\, dx$$$을(를) 구하시오.

$$$u=x - 5$$$라 하자.

그러면 $$$du=\left(x - 5\right)^{\prime }dx = 1 dx$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dx = du$$$임을 얻습니다.

따라서,

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

Expand the expression:

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

각 항별로 적분하십시오:

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

멱법칙($$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=15$$$에 적용합니다:

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

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=10$$$와 $$$f{\left(u \right)} = u^{14}$$$에 적용하세요:

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

멱법칙($$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=14$$$에 적용합니다:

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

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=25$$$와 $$$f{\left(u \right)} = u^{13}$$$에 적용하세요:

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

멱법칙($$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=13$$$에 적용합니다:

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

다음 $$$u=x - 5$$$을 기억하라:

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

따라서,

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

간단히 하시오:

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

적분 상수를 추가하세요:

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

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