Integral of $$$\frac{\sigma_{1}^{2} \sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}$$$ with respect to $$$\sigma_{1}$$$

Find $$$\int \frac{\sigma_{1}^{2} \sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}\, d\sigma_{1}$$$.

Apply the constant multiple rule $$$\int c f{\left(\sigma_{1} \right)}\, d\sigma_{1} = c \int f{\left(\sigma_{1} \right)}\, d\sigma_{1}$$$ with $$$c=\frac{\sigma_{2}^{2} \sigma_{3}}{\sigma_{4}}$$$ and $$$f{\left(\sigma_{1} \right)} = \sigma_{1}^{2}$$$:

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

Apply the power rule $$$\int \sigma_{1}^{n}\, d\sigma_{1} = \frac{\sigma_{1}^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

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

Therefore,

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

Add the constant of integration:

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

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