Integral of $$$\sqrt{u}$$$

Find $$$\int \sqrt{u}\, du$$$.

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

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

Therefore,

$$\int{\sqrt{u} d u} = \frac{2 u^{\frac{3}{2}}}{3}$$

Add the constant of integration:

$$\int{\sqrt{u} d u} = \frac{2 u^{\frac{3}{2}}}{3}+C$$

$$$\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3} + C$$$A