$$$\sigma_{1}$$$에 대한 $$$\frac{\sigma_{1}^{2} \sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}$$$의 적분

$$$\int \frac{\sigma_{1}^{2} \sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}\, d\sigma_{1}$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(\sigma_{1} \right)}\, d\sigma_{1} = c \int f{\left(\sigma_{1} \right)}\, d\sigma_{1}$$$을 $$$c=\frac{\sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}$$$와 $$$f{\left(\sigma_{1} \right)} = \sigma_{1}^{2}$$$에 적용하세요:

$${\color{red}{\int{\frac{\sigma_{1}^{2} \sigma_{2}^{2} \sigma_{3}}{\sigma_{4}} d \sigma_{1}}}} = {\color{red}{\frac{\sigma_{2}^{2} \sigma_{3} \int{\sigma_{1}^{2} d \sigma_{1}}}{\sigma_{4}}}}$$

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

$$\frac{\sigma_{2}^{2} \sigma_{3} {\color{red}{\int{\sigma_{1}^{2} d \sigma_{1}}}}}{\sigma_{4}}=\frac{\sigma_{2}^{2} \sigma_{3} {\color{red}{\frac{\sigma_{1}^{1 + 2}}{1 + 2}}}}{\sigma_{4}}=\frac{\sigma_{2}^{2} \sigma_{3} {\color{red}{\left(\frac{\sigma_{1}^{3}}{3}\right)}}}{\sigma_{4}}$$

따라서,

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

적분 상수를 추가하세요:

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

$$$\int \frac{\sigma_{1}^{2} \sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}\, d\sigma_{1} = \frac{\sigma_{1}^{3} \sigma_{2}^{2} \sigma_{3}}{3 \sigma_{4}} + C$$$A