$$$\frac{\sigma_{1}^{2} \sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}$$$ 关于$$$\sigma_{1}$$$的积分

求$$$\int \frac{\sigma_{1}^{2} \sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}\, d\sigma_{1}$$$。

对 $$$c=\frac{\sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}$$$ 和 $$$f{\left(\sigma_{1} \right)} = \sigma_{1}^{2}$$$ 应用常数倍法则 $$$\int c f{\left(\sigma_{1} \right)}\, d\sigma_{1} = c \int f{\left(\sigma_{1} \right)}\, d\sigma_{1}$$$：

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

应用幂法则 $$$\int \sigma_{1}^{n}\, d\sigma_{1} = \frac{\sigma_{1}^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=2$$$：

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

因此，

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

加上积分常数：

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

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