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