$$$\frac{1}{2 x^{3}}$$$의 적분

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

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=\frac{1}{2}$$$와 $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$에 적용하세요:

$${\color{red}{\int{\frac{1}{2 x^{3}} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{x^{3}} d x}}{2}\right)}}$$

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

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

따라서,

$$\int{\frac{1}{2 x^{3}} d x} = - \frac{1}{4 x^{2}}$$

적분 상수를 추가하세요:

$$\int{\frac{1}{2 x^{3}} d x} = - \frac{1}{4 x^{2}}+C$$

$$$\int \frac{1}{2 x^{3}}\, dx = - \frac{1}{4 x^{2}} + C$$$A