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Can't figure out this integral

  1. Jun 20, 2007 #1
    I've been trying to figure this out for a while, but I can't find the right substitution:

    [tex]\int \sqrt{\frac{x-1}{x(x+1)}} dx [/tex]

    Anyone able to help me out? It's driving me crazy.
  2. jcsd
  3. Jun 20, 2007 #2
    Well, Wolfram seems to think that it reduces to the sum of an elliptic integral of the first kind and second kind, furthermore involving complex numbers.
  4. Jun 20, 2007 #3
    Oh, gee, I copied it wrong. How embarrassing. My apologies. How about

    [tex]\int \frac{\sqrt{\frac{x-1}{x+1}}}{x^2} dx [/tex]

    Wolfram thinks it's arctan-ny.
  5. Jun 20, 2007 #4
    Make the substitution [tex]x = sin^2h(t)[/tex], and the integral reduces to:

    [tex]2\int \sqrt{sin^2h(t) - 1}\ dt[/tex]

    Edit: just saw your post. Oops.
  6. Jun 20, 2007 #5

    D H

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    You get an ugly mess if you type that into Wolfram's Integrator directly. On the other hand, simplifying the radical leads to the much simpler form
    [tex]\int\frac{\sqrt{x^2-1}}{x^2(x+1)}dx =
  7. Jun 20, 2007 #6
    In that case, make the substitution [tex]u^2 =\frac{x-1}{x+1}\ [/tex].

    So the integral becomes:

    [tex]\int \frac{4u^2 du}{(1+u^2)^2}\[/tex]

    Looks like a further substitution of u = tan(t) or u = sinh(a) will do the trick. Or partial fractions.

    By partial fractions, the integrand is: [tex]\frac{4}{1+u^2}\ - \frac{4}{(1+u^2)^2}\[/tex]
    Last edited: Jun 20, 2007
  8. Jun 20, 2007 #7
    I'm having trouble with the algebra manipulating that substitution to get it nicely as

    [tex]\int \frac{4u^2}{(1+u^2)^2} du[/tex]

    Would it be too much trouble to explain how you did that? Of course from there, it's nice.
  9. Jun 21, 2007 #8
    [tex]u^2 = \frac{x-1}{x+1}\ [/tex]
    [tex] = 1 - \frac{2}{x+1}\ [/tex]

    Solving for (x + 1) gives:

    [tex] x + 1 = \frac{2}{1 - u^2}\ [/tex] --- (1)

    So, [tex](x + 1)^2 = \frac{4}{(1 - u^2)^2}\ [/tex] --- (2)

    Also, [tex]2u du = \frac{2 dx}{(x+1)^2}\ [/tex]

    ^ You got that part right? So:

    [tex]dx = u(x + 1)^2 du[/tex]

    From (2),

    [tex]dx = \frac{4u du}{(1 - u^2)^2}\ [/tex]

    So, the expression inside the square root of your original integral becomes u^2, so the whole square root becomes u. From (1):

    [tex] x = \frac{2 - (1 - u^2)}{1 - u^2}\ [/tex]

    [tex] = \frac{1 + u^2}{1 - u^2}\ [/tex]

    Therefore, [tex]x^2 = \frac{(1 + u^2)^2}{(1 - u^2)^2}\ [/tex]

    So, the whole integral becomes:

    [tex] \int \frac{ \frac{4u^2 du}{(1 - u^2)^2} }{ \frac{(1 + u^2)^2}{(1 - u^2)^2}\ } \ [/tex]

    = [tex] \int \frac{4u^2 du}{(1 + u^2)^2}\ [/tex]
  10. Jun 21, 2007 #9
    Ahhh I get it now. That was very clear. :smile:

    Thanks very much for your help. Now I can sleep contentedly!
  11. Jun 21, 2007 #10
    well i tried it by substituting the following:

    [tex]x=\tan^2\t[/tex] and then continued. It simplifies very well.
  12. Jun 21, 2007 #11
    I never get my latex coding right.
  13. Jun 21, 2007 #12
    well isnt the backslash to be for space,

  14. Jun 21, 2007 #13
    now i got it all wrong
  15. Jun 21, 2007 #14
    Now it is fine, sorry for my disobedience with so many posts.
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