$$$\frac{1}{2 \sqrt{x}}$$$의 적분

$$$\int \frac{1}{2 \sqrt{x}}\, dx$$$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 $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$에 적용하세요:

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

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

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

따라서,

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

적분 상수를 추가하세요:

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

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