$$$x \left(20 x - 10\right) + \sqrt{3}$$$의 적분

$$$\int \left(x \left(20 x - 10\right) + \sqrt{3}\right)\, dx$$$을(를) 구하시오.

각 항별로 적분하십시오:

$${\color{red}{\int{\left(x \left(20 x - 10\right) + \sqrt{3}\right)d x}}} = {\color{red}{\left(\int{\sqrt{3} d x} + \int{x \left(20 x - 10\right) d x}\right)}}$$

상수 법칙 $$$\int c\, dx = c x$$$을 $$$c=\sqrt{3}$$$에 적용하십시오:

$$\int{x \left(20 x - 10\right) d x} + {\color{red}{\int{\sqrt{3} d x}}} = \int{x \left(20 x - 10\right) d x} + {\color{red}{\sqrt{3} x}}$$

피적분함수를 단순화하세요.:

$$\sqrt{3} x + {\color{red}{\int{x \left(20 x - 10\right) d x}}} = \sqrt{3} x + {\color{red}{\int{10 x \left(2 x - 1\right) d x}}}$$

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=10$$$와 $$$f{\left(x \right)} = x \left(2 x - 1\right)$$$에 적용하세요:

$$\sqrt{3} x + {\color{red}{\int{10 x \left(2 x - 1\right) d x}}} = \sqrt{3} x + {\color{red}{\left(10 \int{x \left(2 x - 1\right) d x}\right)}}$$

Expand the expression:

$$\sqrt{3} x + 10 {\color{red}{\int{x \left(2 x - 1\right) d x}}} = \sqrt{3} x + 10 {\color{red}{\int{\left(2 x^{2} - x\right)d x}}}$$

각 항별로 적분하십시오:

$$\sqrt{3} x + 10 {\color{red}{\int{\left(2 x^{2} - x\right)d x}}} = \sqrt{3} x + 10 {\color{red}{\left(- \int{x d x} + \int{2 x^{2} d x}\right)}}$$

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

$$\sqrt{3} x + 10 \int{2 x^{2} d x} - 10 {\color{red}{\int{x d x}}}=\sqrt{3} x + 10 \int{2 x^{2} d x} - 10 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\sqrt{3} x + 10 \int{2 x^{2} d x} - 10 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

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

$$- 5 x^{2} + \sqrt{3} x + 10 {\color{red}{\int{2 x^{2} d x}}} = - 5 x^{2} + \sqrt{3} x + 10 {\color{red}{\left(2 \int{x^{2} d x}\right)}}$$

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

$$- 5 x^{2} + \sqrt{3} x + 20 {\color{red}{\int{x^{2} d x}}}=- 5 x^{2} + \sqrt{3} x + 20 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- 5 x^{2} + \sqrt{3} x + 20 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

따라서,

$$\int{\left(x \left(20 x - 10\right) + \sqrt{3}\right)d x} = \frac{20 x^{3}}{3} - 5 x^{2} + \sqrt{3} x$$

간단히 하시오:

$$\int{\left(x \left(20 x - 10\right) + \sqrt{3}\right)d x} = \frac{x \left(20 x^{2} - 15 x + 3 \sqrt{3}\right)}{3}$$

적분 상수를 추가하세요:

$$\int{\left(x \left(20 x - 10\right) + \sqrt{3}\right)d x} = \frac{x \left(20 x^{2} - 15 x + 3 \sqrt{3}\right)}{3}+C$$

$$$\int \left(x \left(20 x - 10\right) + \sqrt{3}\right)\, dx = \frac{x \left(20 x^{2} - 15 x + 3 \sqrt{3}\right)}{3} + C$$$A