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

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

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

Therefore,

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

Add the constant of integration:

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

Answer: $$$\int{\frac{1}{\sqrt{x}} d x}=2 \sqrt{x}+C$$$