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

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

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

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

Therefore,

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

Add the constant of integration:

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

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