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Homework Help: Integration of inverse root

  1. Feb 11, 2006 #1
    I am taking calc 2 and I think we just finished up all the different ways of integrating, yet I cant figure this seemingly very simple one out. :grumpy: Any help is greatly appriciated.o:)

    [tex]\int\sqrt\frac{1}{x}dx[/tex]
     
  2. jcsd
  3. Feb 11, 2006 #2

    VietDao29

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    Homework Helper

    Hint: You know this
    [tex]\int x ^ {\alpha} dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C, \quad \alpha \neq -1[/tex], right?
    [tex]\int \sqrt{\frac{1}{x}} dx = \int \frac{dx}{\sqrt{x}} = \int x ^ {-\frac{1}{2}} dx = ?[/tex]
    Can you go from here? :)
     
  4. Feb 11, 2006 #3
    sure, but when I try that, the answer does not work. That yeilds:
    [tex]2\sqrt{x}[/tex]

    I know the area under the curve already, and when I use [tex]2\sqrt{x}[/tex] in the definate integral, it does not give the correct area.
     
  5. Feb 11, 2006 #4
    I think the answer should be:
    [tex]2x\sqrt{1/x}[/tex]
    but I am not sure how to get there.
     
  6. Feb 11, 2006 #5
    Ummm... but that is equivalent to [tex]2\sqrt{x}[/tex]
     
  7. Feb 11, 2006 #6

    VietDao29

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    Homework Helper

    May you tell me the whole problem? To me, that's correct.
    Just a small error there: You forgot to add the constant of integration (i.e, + C) into your result...
     
  8. Feb 11, 2006 #7
    although it is really quite confusing becuase when I use my calculator to integrate [tex]\int\sqrt{\frac{1}{x}}[/tex], it gives the [tex]2x\sqrt{\frac{1}{x}}[/tex]. Yet, when I use the calculator to differentiate [tex]2x\sqrt{\frac{1}{x}}[/tex], it does not give me [tex]\int\sqrt{\frac{1}{x}}[/tex]. Unless, it gives it in a more complicated form that I have yet to work back to the original.
     
  9. Feb 11, 2006 #8
    can you explain? Could be that my algebra is lacking.
     
  10. Feb 11, 2006 #9
    What are you talking about? Could you restate this a bit more clearly.
     
  11. Feb 11, 2006 #10
    the square root of 1/x is the square root of 1 divided by the square root of x, which is equal to 1 over the square root of x, and x divided by the square root of x is the square root of x, multiply that times 2 and you get 2 times the square root of x.
     
  12. Feb 11, 2006 #11

    VietDao29

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    Homework Helper

    Okay, one suggestion, though. Please don't rely heavily on calculators, and you can try do it by pen and paper, right?
    [tex]2x \sqrt{\frac{1}{x}} = 2(\sqrt{x}) ^ 2 \frac{1}{\sqrt{x}} = 2 \sqrt{x}[/tex]. Assuming that x >= 0. Can you get this? :)
     
  13. Feb 11, 2006 #12
    okay, nevermind,

    [tex]2\sqrt{x}[/tex]

    does work in the definate integral... and so it is the correct solution. Maybe it is time for me to go to bed.

    Thanks for the help, I really appriciate it!
     
  14. Feb 11, 2006 #13
    bit drunk

    bit drunk but
    [tex]\int\sqrt\frac{1}{x}dx[/tex]
    equals
    [tex] 2 \sqrt{\frac{1}{x}}x [/tex]
     
  15. Nov 15, 2009 #14
    by the way, what's the proof for this inverse root? I swear I knew it at one time.

    [tex]\int \frac{dx}{(x^2 + a^2)^{3/2}}=\frac{x}{a^2 \sqrt{x^2 + a^2}}[/tex]
     
  16. Nov 15, 2009 #15

    Mark44

    Staff: Mentor

    I can't believe this thread went on for as long as it did without the OP realizing that the integrand could be greatly simplified.
    [tex]\sqrt{\frac{1}{x}}~=~\frac{\sqrt{1}}{\sqrt{x}}~=~\frac{1}{x^{1/2}}~=~x^{-1/2}[/tex]

    So the antiderivative requires nothing more than an application of the power rule.
     
  17. Nov 15, 2009 #16
    wow.

    3 years later, and yes, I think I've finally figured it out.

    So, I am sitting at my desk trying to wrap my mind around the Kroneker Delta and the Levi-Civita Symbol. I do a google search for info, and a thread on my old friend, the Physics Forums pops up. I click the link, read it, (it did not help), I move on. Now, a few hours later I get an email about a response to a thread I posted three years ago.

    I am guessing PF recognized me and made my account active again.

    Anyhow... yeah, I got that one figured out. Funny how it seems so simple once you "speak" math. But when your just learning, the leap from x/root(x) to root(x) can be evasive.
     
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