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Antiderivative of 1/(x^2 - 1)

  1. Mar 1, 2005 #1
    Do you guys have any tips on how to get the antiderivative of

    1 / (x^2 - 1)?

    Thanks.
     
  2. jcsd
  3. Mar 1, 2005 #2

    dextercioby

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    Yes,simple fractions would be the most direct and elegant way.

    Daniel.
     
  4. Mar 1, 2005 #3
    [tex]\int \frac{1}{(x-1)(x+1)}dx = \int {\frac{1}{2(x-1)}} - {\frac{1}{2(x+1)}}dx[/tex]
     
    Last edited: Mar 2, 2005
  5. Mar 1, 2005 #4

    dextercioby

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    That was my first guess.If you wanna do something fancy,how about computing the antiderivative of argument tangent hyperbolicus...?

    Daniel.
     
  6. Mar 1, 2005 #5

    Hurkyl

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    You forgot the dx's. :tongue2:
     
  7. Mar 1, 2005 #6
    Thanks.

    Ans is therefore,

    (1/2) ln(x-1) - (1/2) ln(x+1)
     
  8. Mar 1, 2005 #7

    dextercioby

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    Plus a constant.And you can write it as a natural logarithm of a fraction to which you incorporate the constant.

    Daniel.
     
  9. Mar 1, 2005 #8
    Got it. Thanks.
     
  10. Jan 18, 2012 #9
    and if it is bounded at -3 and -2
     
  11. Jan 18, 2012 #10
    The thread is seven years old, why revive it? If you have a homework question please make a new thread instead of resurrecting old ones.
     
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