Rewrite $\frac{1}{\sqrt{x}}$ 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 $\frac{x^{\frac{1}{2}}}{\frac{1}{2}}$. Simplify and convert from exponential to radical form.
$$\sqrt{2}\sqrt{x}$$
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.
