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What are the steps required to solve the following?

$\int \sqrt{\frac{11}{x}}\,\mathrm{d}x$

I'm not looking for anyone to do my homework. I usually have no problem figuring these things out -- using Wolfram Alpha step-by-step if absolutely necessary -- but for some reason this seemingly simple problem has me stumped.

Below are the Wolfram Alpha step-by-step instructions for doing this. I get lost on the part where they do the u substitutions.

Any help is greatly appreciated.

enter image description here

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  • $\begingroup$ Remove a constant, and then write a power of $x$. $\endgroup$ – user61527 Jul 15 '14 at 1:45
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There's no need for a substitution, and that only complicates matters and obscures what's important. Write

$$\sqrt{\frac{11}{x}} = \frac{\sqrt{11}}{\sqrt x} = \sqrt{11} x^{-1/2}$$

Now integrate:

\begin{align*} \int \sqrt{11} x^{-1/2} dx &= \sqrt{11} \int x^{-1/2} dx \\ &= \sqrt{11} \frac{x^{-1/2 + 1}}{-1/2 + 1} + C \\ &= \sqrt{11} \frac{x^{1/2}}{1/2} + C \\ &= 2 \sqrt{11} x^{1/2} + C \\ &= 2 \sqrt{11 x} + C \end{align*}

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  • $\begingroup$ Ah yes. I couldn't see the forest for the trees. Thank you very much. $\endgroup$ – WXB13 Jul 15 '14 at 1:59
  • $\begingroup$ You're very welcome. You should always look to pull constants out, and deal with single powers of $x$. $\endgroup$ – user61527 Jul 15 '14 at 2:07
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Suggestion $\frac{1}{\sqrt{x}} = x^{-1/2}$.

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  • $\begingroup$ Thanks. I can't believe that I didn't figure that out on my own. $\endgroup$ – WXB13 Jul 15 '14 at 2:02
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Hint:

Remove the constant. Then note that $\sqrt\frac{1}{x} = \frac{1}{\sqrt{x}} = x^{\frac{-1}{2}}$.

You should be able to go from here simply applying the Fundamental Theorem of Calculus. The derivative of what will give you $x^{\frac{-1}{2}}$?

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  • $\begingroup$ Yes, it was simple as I guessed that it was. I think I was thrown by Wolfram Aplha's convoluted solution (and result). Thanks for the help. $\endgroup$ – WXB13 Jul 15 '14 at 2:01
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    $\begingroup$ No problem @GaryWhite. Unfortunately, WolframAlpha isn't human, so the output will often be a bit clunky/unintuitive. I'm glad I could help! $\endgroup$ – Kaj Hansen Jul 15 '14 at 2:03

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