$$$\frac{1}{n^{\frac{3}{2}}}$$$의 적분

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

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

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

따라서,

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

적분 상수를 추가하세요:

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

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