$$$\frac{1}{t^{23}}$$$의 적분

$$$\int \frac{1}{t^{23}}\, dt$$$을(를) 구하시오.

멱법칙($$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=-23$$$에 적용합니다:

$${\color{red}{\int{\frac{1}{t^{23}} d t}}}={\color{red}{\int{t^{-23} d t}}}={\color{red}{\frac{t^{-23 + 1}}{-23 + 1}}}={\color{red}{\left(- \frac{t^{-22}}{22}\right)}}={\color{red}{\left(- \frac{1}{22 t^{22}}\right)}}$$

따라서,

$$\int{\frac{1}{t^{23}} d t} = - \frac{1}{22 t^{22}}$$

적분 상수를 추가하세요:

$$\int{\frac{1}{t^{23}} d t} = - \frac{1}{22 t^{22}}+C$$

$$$\int \frac{1}{t^{23}}\, dt = - \frac{1}{22 t^{22}} + C$$$A