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What is the integral of sqrt(x^2-1)/x dx?

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the integral of sqrt(x^2-1)/x dx?

    2. Relevant equations

    3. The attempt at a solution

    ∫ √(x^2 - 1) dx / x

    let x = sec u: u = sec^-1(x) and tan u = √(x^2 - 1)

    dx = sec u tan u du

    now the integral becomes

    ∫ √sec^2(u) - 1) sec u tan u du / sec u

    = ∫ tan u tan u du

    = ∫ tan^2(u) du

    = ∫ (sec^2(u) - 1) du

    = ∫ (sec^2(u) du - ∫ du

    = tan u - u + c

    back substitute u = sec^-1(x) and tan u = √(x^2 - 1)

    = √(x^2 - 1) - sec^-1(x) + c

    However, the correct answer should be arccot(sqrt(x^2-1))+sqrt(x^2-1) + c. Can someone help me find what went wrong?
  2. jcsd
  3. Jan 18, 2009 #2
    Maple says that this integral is
  4. Jan 18, 2009 #3
    Well, arctan(1/(x^2-1)) is the same as arccot(sqrt(x^2-1)). I just don't know the mistake I did when showing my steps.
  5. Jan 18, 2009 #4
    funzone36, what you initially had was more or less correct. you can check by writing sec^-1(x) as arccos(1/x) and then differentiating.

    The answer that you said was supposedly correct involving arccot is gotten by substituting x = csc(u) and turns out to be a slightly neater approach.
  6. Jan 18, 2009 #5


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    Homework Helper

    arcsec(x)=arctan(sqrt(x^2-1)). Draw a right triangle with hypotenuse x and leg 1. t=arcsec(x) is the angle between those two. What's tan(t)?
  7. Jan 18, 2009 #6
    Thanks everywhere. My steps were actually correct. I learned that I should draw diagrams next time I do these problems.
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