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

Find $$$\int \sqrt{x}\, dx$$$.

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \sqrt{x}\, dx = \frac{2 x^{\frac{3}{2}}}{3} + C$$$A