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Evaluate
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Differentiate w.r.t. x
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2\sqrt{x}
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.
2\sqrt{x}+С
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.