$$$\frac{1}{x^{\frac{3}{2}}}$$$의 적분

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

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

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

따라서,

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

적분 상수를 추가하세요:

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

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