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Integral of 2/(x^2-1)dx HELP

  1. Oct 30, 2011 #1
    Integral of 2/(x^2-1)dx HELP!!!

    1. The problem statement, all variables and given/known data

    Evaluate: ∫2/(x2-1)dx

    2. Relevant equations
    Inverse trig functions arent working.
    ∫1/x+1dx=ln(1+x)


    3. The attempt at a solution

    My tutor and i both attemped the solution many ways, and are still at a stand still.
    According to my calculator and wolfram alpha, the answer is -ln(x+1/x-1)
    The closest i have gotten is 2∫[itex]1/((x+1)(x-1))[/itex]dx
     
  2. jcsd
  3. Oct 30, 2011 #2

    Dick

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    Re: Integral of 2/(x^2-1)dx HELP!!!

    Did you try using partial fractions?
     
  4. Oct 30, 2011 #3
    Re: Integral of 2/(x^2-1)dx HELP!!!

    if you multiply by -1 I think you can use tanh^-1
     
  5. Oct 30, 2011 #4
    Re: Integral of 2/(x^2-1)dx HELP!!!

    Thanks! partial fractions worked! didn't even think about that. the solution came out to be -ln(x+1)+ln(x-1), which is apparently equal to -ln((x+1)/(x-1))
     
  6. Oct 30, 2011 #5
    Re: Integral of 2/(x^2-1)dx HELP!!!

    and also apparently equal to tanh^-1
     
  7. Oct 30, 2011 #6
    Re: Integral of 2/(x^2-1)dx HELP!!!

    The full question is to find the integral over infinity and 2, so i have to use limits. im getting an undefined answer since ln of infinity minus ln infinity is undefined. does that sound right?
     
  8. Oct 30, 2011 #7

    Dick

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    Re: Integral of 2/(x^2-1)dx HELP!!!

    Not right. You should really be writing things like the integral of 1/(1+x) as log(|1+x|). Since it's an improper integral, to work out the infinity part of your limit you need to think about the limit as x->infinity of log(|1+x|/|1-x|). What's that?
     
  9. Oct 30, 2011 #8
    Re: Integral of 2/(x^2-1)dx HELP!!!

    should be zero right? Not used to using log to write things, its just something our teacher never really told us to do. makes sense though
     
  10. Oct 30, 2011 #9

    Mark44

    Staff: Mentor

    Re: Integral of 2/(x^2-1)dx HELP!!!

    Although my choice would be partial fractions decomposition, a trig substitution will also work here, with secθ = x, secθtanθdθ = dx. The resulting integral is
    [tex]2\int csc\theta d\theta[/tex]
     
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