$$$\frac{1}{\sqrt{t}}$$$의 적분

$$$\int \frac{1}{\sqrt{t}}\, dt$$$을(를) 구하시오.

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

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

따라서,

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

적분 상수를 추가하세요:

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

$$$\int \frac{1}{\sqrt{t}}\, dt = 2 \sqrt{t} + C$$$A