$$$\frac{1}{\sqrt[3]{x}}$$$의 적분

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

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

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

따라서,

$$\int{\frac{1}{\sqrt[3]{x}} d x} = \frac{3 x^{\frac{2}{3}}}{2}$$

적분 상수를 추가하세요:

$$\int{\frac{1}{\sqrt[3]{x}} d x} = \frac{3 x^{\frac{2}{3}}}{2}+C$$

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