Integral of $$$\frac{51 n}{100}$$$

Find $$$\int \frac{51 n}{100}\, dn$$$.

Apply the constant multiple rule $$$\int c f{\left(n \right)}\, dn = c \int f{\left(n \right)}\, dn$$$ with $$$c=\frac{51}{100}$$$ and $$$f{\left(n \right)} = n$$$:

$${\color{red}{\int{\frac{51 n}{100} d n}}} = {\color{red}{\left(\frac{51 \int{n d n}}{100}\right)}}$$

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

$$\frac{51 {\color{red}{\int{n d n}}}}{100}=\frac{51 {\color{red}{\frac{n^{1 + 1}}{1 + 1}}}}{100}=\frac{51 {\color{red}{\left(\frac{n^{2}}{2}\right)}}}{100}$$

Therefore,

$$\int{\frac{51 n}{100} d n} = \frac{51 n^{2}}{200}$$

Add the constant of integration:

$$\int{\frac{51 n}{100} d n} = \frac{51 n^{2}}{200}+C$$

$$$\int \frac{51 n}{100}\, dn = \frac{51 n^{2}}{200} + C$$$A