$$$\frac{1}{2 u^{2}}$$$의 적분

$$$\int \frac{1}{2 u^{2}}\, du$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=\frac{1}{2}$$$와 $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$에 적용하세요:

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

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

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

따라서,

$$\int{\frac{1}{2 u^{2}} d u} = - \frac{1}{2 u}$$

적분 상수를 추가하세요:

$$\int{\frac{1}{2 u^{2}} d u} = - \frac{1}{2 u}+C$$

$$$\int \frac{1}{2 u^{2}}\, du = - \frac{1}{2 u} + C$$$A