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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.

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2 Answers 2

up vote 1 down vote accepted

Think of $\frac{1}{\sqrt{u}}$ as $u^{-\frac{1}{2}}$. Now you can just apply the power rule (i.e. increase the exponent rule by 1 and divide by the new exponent). This gives $$ \begin{align*} \int \frac{1}{\sqrt{u}} \, du &= \int u^{-\frac{1}{2}} \, du\\ &= \frac{u^\frac{1}{2}}{\frac{1}{2}} + C\\ &= 2 \sqrt{u} + C. \end{align*} $$

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Ah nuts. I was getting confused with the derivative rules. I mixed them up. Duh moment. Thanks. I'll accept when I can. :) –  asdf Oct 8 '11 at 1:57

Check it with $\frac{d}{du} (\sqrt{u}+C)$.

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That derivative evaluates to $0$. :) –  J. M. Oct 11 '11 at 3:58

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