Integral of $$$2 n^{\frac{3}{2}}$$$

Find $$$\int 2 n^{\frac{3}{2}}\, dn$$$.

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

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

Apply the power rule $$$\int n^{n}\, dn = \frac{n^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{3}{2}$$$:

$$2 {\color{red}{\int{n^{\frac{3}{2}} d n}}}=2 {\color{red}{\frac{n^{1 + \frac{3}{2}}}{1 + \frac{3}{2}}}}=2 {\color{red}{\left(\frac{2 n^{\frac{5}{2}}}{5}\right)}}$$

Therefore,

$$\int{2 n^{\frac{3}{2}} d n} = \frac{4 n^{\frac{5}{2}}}{5}$$

Add the constant of integration:

$$\int{2 n^{\frac{3}{2}} d n} = \frac{4 n^{\frac{5}{2}}}{5}+C$$

$$$\int 2 n^{\frac{3}{2}}\, dn = \frac{4 n^{\frac{5}{2}}}{5} + C$$$A