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İntegral of Rational Function

  1. Feb 25, 2017 #1
    1. The problem statement, all variables and given/known data
    ##∫\frac {dx}{(x^2-1)^2}##

    2. Relevant equations


    3. The attempt at a solution
    I tried to divide ##\frac {1} {(x^2-1)^2}## as ##\frac {Ax+B} {(x^2-1)} +\frac {Cx^3+Dx^2+Ex+F} {(x^2-1)^2}##

    but this looks so complex..I dont know how to do ? Maybe I can just left at simpler terms.But how can I decide it ?
     
  2. jcsd
  3. Feb 25, 2017 #2

    PeroK

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    One trick is to just use a single label for a polynomial:

    ##\frac {1} {(x^2-1)^2} = \frac {A(x)} {(x^2-1)} +\frac {B(x)} {(x^2-1)^2}##

    And see what that leads to.
     
  4. Feb 25, 2017 #3
    Is this specifically meant to be done by partial fractions? I would try a trig substitution first. Didn't do it so I don't know what happens.
     
  5. Feb 25, 2017 #4

    Ray Vickson

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    Doesn't it lead to ##A(x) =0## and ##B(x) =1?##

    However, the OP could certainly get
    $$\frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1},$$
    and thus get
    $$\frac{1}{(x^2-1)^2} = \frac{A^2}{(x-1)^2}+ \frac{2 A B}{(x-1)(x+1)} + \frac{B^2}{(x+1)^2}, $$
    then do something with that.
     
  6. Feb 25, 2017 #5

    Mark44

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    If the integral is meant to be evaluated using partial fractions, I would decompose the integrand like this:
    ##\frac 1 {(x^2-1)^2} = \frac 1 {(x - 1)^2(x + 1)^2} = \frac A {x - 1} +\frac B {(x - 1)^2}+ \frac C {x + 1} + \frac D {(x + 1)^2} ##
     
  7. Feb 25, 2017 #6

    PeroK

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    There is another solution!
     
  8. Feb 25, 2017 #7
    I get ##A(x)=0##
    ##B(x)-2A(x)=0##
    ##A(x)-B(x)=1##
    whats the other solution ?

    Is it make things easier ? (I am not sure..)
    can lead us yes but too long...
     
  9. Feb 25, 2017 #8

    PeroK

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    The other solution is fairly obvious, but it doesn't lead to a great improvement. You shouldn't have been so hasty to dismiss @Mark44's suggestion. I think your definition of what is "too complicated" is no longer appropriate for the level of mathematics you are dealing with. You shouldn't be baulking at only four coefficients.
     
  10. Feb 25, 2017 #9
    Its not too hard to find a b c d but after that.We will left another ##(x-1)^2## and ##(x+1)^2## so I have to do all stuff again..Fine I ll try
     
  11. Feb 25, 2017 #10

    Ray Vickson

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    Not too long at all. Sometimes things need considerable work, and there is no way to avoid doing it. Sometimes there are just no shortcuts.
     
  12. Feb 25, 2017 #11
    The answer is ##\frac {1} {4} ln (\frac {x+1} {x-1}) - \frac{x} {2(x^2-1)}+C##

    I ll find soon
     
  13. Feb 25, 2017 #12
    yeah I know..But they shouldnt made such long questions
     
  14. Feb 25, 2017 #13

    Mark44

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    Real life isn't always about short questions.

    You never told us whether you had to use a specific technique (such as by using partial fraction decomposition). Another approach that was already mentioned was trig substitution.
     
  15. Feb 25, 2017 #14
    You are right.The question is in the partial fraction section
     
  16. Feb 25, 2017 #15

    PeroK

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    "Nature laughs at the difficulties of integration." - Laplace
     
  17. Feb 25, 2017 #16
    I tried every method that we thought but nothing came up
     
  18. Feb 25, 2017 #17

    Ray Vickson

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    What partial fraction representation are you trying to use? Which parts yield "nothing"?
     
  19. Feb 26, 2017 #18
    Its too long to write down here :/.I can take photo maybe..Or I ll ask just my prof more simple
     
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