$$$\left(x + 1\right) \left(x + 2\right)$$$의 적분

$$$\int \left(x + 1\right) \left(x + 2\right)\, dx$$$을(를) 구하시오.

Expand the expression:

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

각 항별로 적분하십시오:

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

상수 법칙 $$$\int c\, dx = c x$$$을 $$$c=2$$$에 적용하십시오:

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

멱법칙($$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=2$$$에 적용합니다:

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

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

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

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

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

따라서,

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

간단히 하시오:

$$\int{\left(x + 1\right) \left(x + 2\right) d x} = \frac{x \left(2 x^{2} + 9 x + 12\right)}{6}$$

적분 상수를 추가하세요:

$$\int{\left(x + 1\right) \left(x + 2\right) d x} = \frac{x \left(2 x^{2} + 9 x + 12\right)}{6}+C$$

$$$\int \left(x + 1\right) \left(x + 2\right)\, dx = \frac{x \left(2 x^{2} + 9 x + 12\right)}{6} + C$$$A