$$$- \sqrt[3]{3} \sqrt[3]{x} - 1$$$의 적분

$$$\int \left(- \sqrt[3]{3} \sqrt[3]{x} - 1\right)\, dx$$$을(를) 구하시오.

각 항별로 적분하십시오:

$${\color{red}{\int{\left(- \sqrt[3]{3} \sqrt[3]{x} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} - \int{\sqrt[3]{3} \sqrt[3]{x} d x}\right)}}$$

상수 법칙 $$$\int c\, dx = c x$$$을 $$$c=1$$$에 적용하십시오:

$$- \int{\sqrt[3]{3} \sqrt[3]{x} d x} - {\color{red}{\int{1 d x}}} = - \int{\sqrt[3]{3} \sqrt[3]{x} d x} - {\color{red}{x}}$$

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

$$- x - {\color{red}{\int{\sqrt[3]{3} \sqrt[3]{x} d x}}} = - x - {\color{red}{\sqrt[3]{3} \int{\sqrt[3]{x} d x}}}$$

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

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

따라서,

$$\int{\left(- \sqrt[3]{3} \sqrt[3]{x} - 1\right)d x} = - \frac{3 \sqrt[3]{3} x^{\frac{4}{3}}}{4} - x$$

적분 상수를 추가하세요:

$$\int{\left(- \sqrt[3]{3} \sqrt[3]{x} - 1\right)d x} = - \frac{3 \sqrt[3]{3} x^{\frac{4}{3}}}{4} - x+C$$

$$$\int \left(- \sqrt[3]{3} \sqrt[3]{x} - 1\right)\, dx = \left(- \frac{3 \sqrt[3]{3} x^{\frac{4}{3}}}{4} - x\right) + C$$$A