$$$- \frac{1}{x^{\frac{2}{3}}}$$$의 적분

$$$\int \left(- \frac{1}{x^{\frac{2}{3}}}\right)\, dx$$$을(를) 구하시오.

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

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

멱법칙($$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=- \frac{2}{3}$$$에 적용합니다:

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

따라서,

$$\int{\left(- \frac{1}{x^{\frac{2}{3}}}\right)d x} = - 3 \sqrt[3]{x}$$

적분 상수를 추가하세요:

$$\int{\left(- \frac{1}{x^{\frac{2}{3}}}\right)d x} = - 3 \sqrt[3]{x}+C$$

$$$\int \left(- \frac{1}{x^{\frac{2}{3}}}\right)\, dx = - 3 \sqrt[3]{x} + C$$$A