Rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$. Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{\frac{1}{2}}\mathrm{d}x$ with $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}$. Simplify.
$$\frac{2x^{\frac{3}{2}}}{3}$$
If $F\left(x\right)$ is an antiderivative of $f\left(x\right)$, then the set of all antiderivatives of $f\left(x\right)$ is given by $F\left(x\right)+C$. Therefore, add the constant of integration $C\in \mathrm{R}$ to the result.
