1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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

    User Avatar
    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

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integration of inverse root
Loading...