Integral of $$$- n + \sigma^{3}$$$ with respect to $$$n$$$

Find $$$\int \left(- n + \sigma^{3}\right)\, dn$$$.

Integrate term by term:

$${\color{red}{\int{\left(- n + \sigma^{3}\right)d n}}} = {\color{red}{\left(- \int{n d n} + \int{\sigma^{3} d n}\right)}}$$

Apply the constant rule $$$\int c\, dn = c n$$$ with $$$c=\sigma^{3}$$$:

$$- \int{n d n} + {\color{red}{\int{\sigma^{3} d n}}} = - \int{n d n} + {\color{red}{n \sigma^{3}}}$$

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

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

Therefore,

$$\int{\left(- n + \sigma^{3}\right)d n} = - \frac{n^{2}}{2} + n \sigma^{3}$$

Simplify:

$$\int{\left(- n + \sigma^{3}\right)d n} = \frac{n \left(- n + 2 \sigma^{3}\right)}{2}$$

Add the constant of integration:

$$\int{\left(- n + \sigma^{3}\right)d n} = \frac{n \left(- n + 2 \sigma^{3}\right)}{2}+C$$

$$$\int \left(- n + \sigma^{3}\right)\, dn = \frac{n \left(- n + 2 \sigma^{3}\right)}{2} + C$$$A