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Integration problem

  1. Feb 17, 2007 #1
    im trying to figure out the integral of 1/(x^2-1).
    i basically have no clue how to do this, since every thing ive tried has resulted in a dead end
    i tried intergration by parts by factoring the denominator but that just results in a more complex integral.
    i also tried tried multiplying both sides of the fraction by 2x and then substituting x^2-1 with u but that doesnt get anywhere either
     
  2. jcsd
  3. Feb 17, 2007 #2

    TD

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    Factor the denominator and split it in two, using partial fraction (decomposition).
     
  4. Feb 17, 2007 #3
    i think this does work by integration by parts

    show us what you did and maybe someone with more knowledge can answer you or point out where you're going wrong.
     
  5. Feb 17, 2007 #4
    alright, what i did when i tried to integrate it by parts is split the denominator
    (1/x-1)*(1/x+1) then i took 1/(x-1) as u and 1/(x+1) as dv. then the new intergral i have to solve is vdu, or the integral of (x-1)^(-2)*ln(x+1) well in this case, ill have to integrate by parts again, but if i choose ln(x+1) as u, then i just get the integral i had before and if i choose (x-1)^(-2) as u, the integral just becomes more complex. did i do sometime wrong?
     
  6. Feb 17, 2007 #5

    TD

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    I really don't see why you'd want to use integration by parts here.
    Try my suggestion, it's obviously the standard approach for this one.
     
  7. Feb 17, 2007 #6
    You can do it by parts and it should be done to make it a little easier and aesthetically pleasing =). You just need to setup a triangle. Draw a triangle first, label each side accordingly.

    After that, you can get your trig functions out of it. Then you can take your original x that you solved for and put it in.

    You need to look for a few things to replace: the dx (which you will solve for x, differentiate that and you get your dx to replace that, then you can plug that back in the end [what x equals]). The denominator part and/or the numerator.
     
    Last edited: Feb 17, 2007
  8. Feb 17, 2007 #7
    TD, i wasn't really taught formally about decomposition but from what im seeing on wikipedia, its basically factoring the denominator, then using substitution right? but if i do that, i just end up with 1/u*(u-2) or 1/u*(u+2), which doesnt help me much. could you maybe show what you mean by decomposition if this wasnt what you meant?
     
  9. Feb 18, 2007 #8
    It looks to me that the best method is to use partial fractions as mentioned.

    [tex] \frac{1}{x^2-1}[/tex]

    Then make x^2-1 into two terms

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

    [tex]\frac{1}{x+1}.\frac{1}{x-1}[/tex]

    Then proceed from there, you don't really need substitution.

    I checked the answer on a maths program and it came up with -atanh(x) hehe, I presume it's equivalent. your answer may well look differrent from this and involve logs.:/
     
    Last edited: Feb 18, 2007
  10. Feb 18, 2007 #9

    arildno

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    Alternatively, use the substitution:
    [tex]x=Tanh(u)\to{dx}=\frac{du}{Cosh^{2}(u)}=du(1-Tanh^{2}(u))=(1-x^{2})du[/tex]
     
  11. Feb 18, 2007 #10

    TD

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    You write the fraction as a sum of fractions with a lineair denominator:

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

    This last expression is easy to integrate.
     
  12. Feb 18, 2007 #11
    oh wow, i feel pretty stupid that i didnt see that before. thanks for the help
     
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