I have a simple equation I typed into WolframAlpha to see what it was. Look what WO gave me:
Indefinite integrals: $ \int \frac{1}{2\sqrt{u}} du = \sqrt{u} + \text{constant}. $ Possible intermediate steps: $ \int \frac{1}{2 \sqrt{u}} du $ Factor out constants: $ \frac 1 2 \int \frac{1}{\sqrt{u}} du $ The integral of $\frac{1}{\sqrt{u}}$ is $2\sqrt{u}$: $ \sqrt{u} + \text{constant} $
Is this so? How can the antiderivative of $\frac{1}{\sqrt{u}}$ be $2 \sqrt{u}$?
Edit [by SN]: Added the steps given by Wolfram|Alpha to the question, and removed the link to the screenshot image.