$$$x^{3} \left(- x - 2\right)$$$의 적분

$$$\int x^{3} \left(- x - 2\right)\, dx$$$을(를) 구하시오.

피적분함수를 단순화하세요.:

$${\color{red}{\int{x^{3} \left(- x - 2\right) d x}}} = {\color{red}{\int{\left(- x^{3} \left(x + 2\right)\right)d x}}}$$

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=-1$$$와 $$$f{\left(x \right)} = x^{3} \left(x + 2\right)$$$에 적용하세요:

$${\color{red}{\int{\left(- x^{3} \left(x + 2\right)\right)d x}}} = {\color{red}{\left(- \int{x^{3} \left(x + 2\right) d x}\right)}}$$

Expand the expression:

$$- {\color{red}{\int{x^{3} \left(x + 2\right) d x}}} = - {\color{red}{\int{\left(x^{4} + 2 x^{3}\right)d x}}}$$

각 항별로 적분하십시오:

$$- {\color{red}{\int{\left(x^{4} + 2 x^{3}\right)d x}}} = - {\color{red}{\left(\int{2 x^{3} d x} + \int{x^{4} d x}\right)}}$$

멱법칙($$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=4$$$에 적용합니다:

$$- \int{2 x^{3} d x} - {\color{red}{\int{x^{4} d x}}}=- \int{2 x^{3} d x} - {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=- \int{2 x^{3} d x} - {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=2$$$와 $$$f{\left(x \right)} = x^{3}$$$에 적용하세요:

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

멱법칙($$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=3$$$에 적용합니다:

$$- \frac{x^{5}}{5} - 2 {\color{red}{\int{x^{3} d x}}}=- \frac{x^{5}}{5} - 2 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- \frac{x^{5}}{5} - 2 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

따라서,

$$\int{x^{3} \left(- x - 2\right) d x} = - \frac{x^{5}}{5} - \frac{x^{4}}{2}$$

간단히 하시오:

$$\int{x^{3} \left(- x - 2\right) d x} = \frac{x^{4} \left(- 2 x - 5\right)}{10}$$

적분 상수를 추가하세요:

$$\int{x^{3} \left(- x - 2\right) d x} = \frac{x^{4} \left(- 2 x - 5\right)}{10}+C$$

$$$\int x^{3} \left(- x - 2\right)\, dx = \frac{x^{4} \left(- 2 x - 5\right)}{10} + C$$$A