$$$\frac{1}{x^{29}}$$$의 적분

$$$\int \frac{1}{x^{29}}\, dx$$$을(를) 구하시오.

멱법칙($$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=-29$$$에 적용합니다:

$${\color{red}{\int{\frac{1}{x^{29}} d x}}}={\color{red}{\int{x^{-29} d x}}}={\color{red}{\frac{x^{-29 + 1}}{-29 + 1}}}={\color{red}{\left(- \frac{x^{-28}}{28}\right)}}={\color{red}{\left(- \frac{1}{28 x^{28}}\right)}}$$

따라서,

$$\int{\frac{1}{x^{29}} d x} = - \frac{1}{28 x^{28}}$$

적분 상수를 추가하세요:

$$\int{\frac{1}{x^{29}} d x} = - \frac{1}{28 x^{28}}+C$$

$$$\int \frac{1}{x^{29}}\, dx = - \frac{1}{28 x^{28}} + C$$$A