Integral of $$$\frac{1}{\sqrt{u}}$$$

Find $$$\int \frac{1}{\sqrt{u}}\, du$$$.

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

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

Therefore,

$$\int{\frac{1}{\sqrt{u}} d u} = 2 \sqrt{u}$$

Add the constant of integration:

$$\int{\frac{1}{\sqrt{u}} d u} = 2 \sqrt{u}+C$$

$$$\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u} + C$$$A