$$$- \frac{1}{2 x^{\frac{3}{2}}}$$$의 적분

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

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

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

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

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

따라서,

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

적분 상수를 추가하세요:

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

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