$$$- \frac{1}{\sqrt{u}}$$$의 적분

$$$\int \left(- \frac{1}{\sqrt{u}}\right)\, du$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=-1$$$와 $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$에 적용하세요:

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

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

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

따라서,

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

적분 상수를 추가하세요:

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

$$$\int \left(- \frac{1}{\sqrt{u}}\right)\, du = - 2 \sqrt{u} + C$$$A