Integral of $$$\frac{2 n}{5} - 1$$$

Find $$$\int \left(\frac{2 n}{5} - 1\right)\, dn$$$.

Integrate term by term:

$${\color{red}{\int{\left(\frac{2 n}{5} - 1\right)d n}}} = {\color{red}{\left(- \int{1 d n} + \int{\frac{2 n}{5} d n}\right)}}$$

Apply the constant rule $$$\int c\, dn = c n$$$ with $$$c=1$$$:

$$\int{\frac{2 n}{5} d n} - {\color{red}{\int{1 d n}}} = \int{\frac{2 n}{5} d n} - {\color{red}{n}}$$

Apply the constant multiple rule $$$\int c f{\left(n \right)}\, dn = c \int f{\left(n \right)}\, dn$$$ with $$$c=\frac{2}{5}$$$ and $$$f{\left(n \right)} = n$$$:

$$- n + {\color{red}{\int{\frac{2 n}{5} d n}}} = - n + {\color{red}{\left(\frac{2 \int{n d n}}{5}\right)}}$$

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

$$- n + \frac{2 {\color{red}{\int{n d n}}}}{5}=- n + \frac{2 {\color{red}{\frac{n^{1 + 1}}{1 + 1}}}}{5}=- n + \frac{2 {\color{red}{\left(\frac{n^{2}}{2}\right)}}}{5}$$

Therefore,

$$\int{\left(\frac{2 n}{5} - 1\right)d n} = \frac{n^{2}}{5} - n$$

Simplify:

$$\int{\left(\frac{2 n}{5} - 1\right)d n} = \frac{n \left(n - 5\right)}{5}$$

Add the constant of integration:

$$\int{\left(\frac{2 n}{5} - 1\right)d n} = \frac{n \left(n - 5\right)}{5}+C$$

$$$\int \left(\frac{2 n}{5} - 1\right)\, dn = \frac{n \left(n - 5\right)}{5} + C$$$A