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

  • Thread starter masonm127
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Integral of 2/(x^2-1)dx HELP!!!

Homework Statement



Evaluate: ∫2/(x2-1)dx

Homework Equations


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


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
 

Answers and Replies

  • #2
Dick
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Did you try using partial fractions?
 
  • #3
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if you multiply by -1 I think you can use tanh^-1
 
  • #4
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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))
 
  • #5
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and also apparently equal to tanh^-1
 
  • #6
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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?
 
  • #7
Dick
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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?
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?
 
  • #8
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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
 
  • #9
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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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