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Seemingly easy integral

  1. Nov 14, 2006 #1
    I came to this integral which I thaught would be easy to solve, but I'm stuck.

    int dx/{x *sqrt{(x ^ 2 )+ k}}
    (Sorry Im trying to get it in Latex)

    I tried substitution:
    [tex]x ^ 2 + k = t[/tex]
    as I remembered from the book (years ago)

    But have ended with this:
    1/2 times
    int dt/{(t-k)sqrt(t)}

    Im not sure if this leads to a solution. Any help would be appreciated
     
    Last edited: Nov 15, 2006
  2. jcsd
  3. Nov 14, 2006 #2
    Try the substitution x=k1/2tan(y)
     
  4. Nov 14, 2006 #3
    Thanks d leet. Sorry I tried in Latex and kept getting most unusual expressions. This is the correct expression.
     
  5. Nov 15, 2006 #4

    dextercioby

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    Try the substitution x=k^{1/2} sinh t.

    Daniel.
     
  6. Nov 18, 2006 #5
    that substitution led me to
    int dt/sinh t
    Any idea how this could be solved?
     
  7. Nov 19, 2006 #6
    It shouldn't lead you to that, can you show the work you did using that substitution?
     
  8. Nov 19, 2006 #7
    Sure if you don't mind my ascii typing :)
    dx = (sqrt)k cosht dt
    I = (sqrt)k*int cosht*dt/((sqrt)k*sinht*(sqrt)(k(sinh^2t + 1))) =

    =(sqrt)k*int cosht*dt/((sqrt)k*sinht*(sqrt)(k*cosh^2t)) =

    ==(sqrt)k*int cosht*dt/(k*sinht*cosht)

    Then after canceling cosht and puting 1/k out, Im left with dt/sinht
     
  9. Nov 19, 2006 #8
    This line is almost impossible to follow, but it doesn't look correct at all. you should have k.5*cosht/(ksinh2 + k).5 inside of the integral.
     
  10. Nov 19, 2006 #9
    I got the same under the root of the denominator. When you factor out k, u got (k(sinh^2 + 1))^.5 But that's only the root. There was an x multilplying that root.
    I think I reduced it properly, but don't know how to solve dt/sinht
     
  11. Nov 19, 2006 #10
    Sorry about that I didn't see the factor of x in the denominator. Well isn't 1/sech(t)= csch(t)? and I think you can look that up in a table of integrals, or I;m sure theres a way to do it, but I'm not sure how do it since I never have before.
     
  12. Nov 21, 2006 #11

    dextercioby

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    Yes, just multiply both the denominator and the numerator by #sinh t# and then use the ch^2 -sh^2 =1 identity.

    Daniel.
     
  13. Nov 21, 2006 #12
    Thanks dextercioby, your first tip was good as I managed to reduce to the above. As I didn't find it in tables, I used an excelent web integrator
    http://integrals.wolfram.com/index.jsp
    The solution is ln(tanh(t/2))
     
  14. Nov 21, 2006 #13

    dextercioby

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    There's also the substitution

    [tex] x^{2}+1 =p^{2} [/tex]

    that leads immediately to a result.

    Daniel.
     
  15. Nov 24, 2006 #14

    benorin

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    www.quickmath.com is an excellent alternative website for indefinite/definite integration, derivatives, plots, solving equations, matricies, etc.
     
  16. Nov 24, 2006 #15

    dextercioby

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    I agree quickmath can perform complicated definite integrals, while Mathematica's web integrator can't.

    Daniel.
     
  17. Nov 24, 2006 #16

    arildno

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    Alternatively, use a u=tanh(t/2) substitution for the fun of it.
     
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