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Integral question

  1. Jan 9, 2008 #1

    i tride to solve it in every way possible
    1.by splitting it into two polinomials
    2.by subtsituting x^4=t
    3.by puttinh x=tan t

    nothing works

    how do i solve it
  2. jcsd
  3. Jan 9, 2008 #2
    As far as I can tell, after some cheating, you're supposed to somehow figure out that [tex] x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1) [/tex]. I'm not sure exactly how one would easily recognize that though (assuming I didn't make a mistake).
  4. Jan 9, 2008 #3


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    Well let's first put it in LaTeX for the benefit of the homework helpers:

    [tex]\int \frac{x^2 +1}{x^4 + 1} dx[/tex]

    How exactly did you cheat Mystic?
  5. Jan 9, 2008 #4
    Wolfram's Integrator.
  6. Jan 9, 2008 #5
    Buy Gelfand's Algebra book and you'll be able to factor almost any polynomial :p
    Last edited: Jan 9, 2008
  7. Jan 9, 2008 #6
    It's somewhat depressing getting recommendations for high school algebra books while sitting in my crappy three-person TA office.

    Anyway, out of curiosity, how would one proceed in getting that factorization of x^4 + 1. I've honestly got no clue, and that would probably be handy information.
  8. Jan 9, 2008 #7
    Gelfand may be intended for HS students, but his books are hard. I just finished Calc 2 and I still struggle on a lot of his problems.



  9. Jan 9, 2008 #8
    Wow, I feel silly now. Thanks.
  10. Jan 9, 2008 #9
    Exactly how I felt before learning a few things in his book :-x
  11. Jan 9, 2008 #10
    After splitting it into 2, and using Mystic's advice ...


    Partial Fraction?
    Last edited: Jan 9, 2008
  12. Jan 9, 2008 #11
    so how do i solve it after this step
  13. Jan 9, 2008 #12
    You have to use Partial Fractions, have you learned that method yet?
  14. Jan 9, 2008 #13
    yes i did but it looks very scary
    will it work?
  15. Jan 9, 2008 #14
    Yes and it's pretty much the only method I can think of. After that, you may want to mess around with completing the square, I did that in order to use arctan.
  16. Jan 12, 2008 #15


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    Gold Member

    Have you done it? Note I think it saves some work if you do not split it as rocophysics suggests but just notice that [tex]\frac{A}{(x^2+\sqrt{2}x+1)}+\frac{B}{(x^2-\sqrt{2}x+1)}[/tex]
    is capable of giving the form [tex]\frac{(x^2 + 1)}{(x^4 + 1)} [/tex] and that part of it is quite simple to get.

    Edit: I mean easy to see or find what A and B must be.
    Last edited: Jan 12, 2008
  17. Jan 12, 2008 #16


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    Since [tex] x^{4}+1 =(x^2 +i) (x^2 -i) [/tex], partial fraction decomposition is simply

    [tex] \frac{x^{2}+1}{x^{4}+1} =\frac{1+i}{2}\frac{1}{x^2 +i} +\frac{1-i}{2}\frac{1}{x^2 -i } [/tex] and then the integration would be easy.
  18. Jan 12, 2008 #17
    The remark of epenguin is correct, you can llok for A and B to make the integrals more easy to solve. However if you notice the following:

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

    And also that:

    [tex]x^4+1 = \left[\left(x+\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2\right] \cdot \left[\left(x-\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2\right][/tex]

    You get two integrals involving arctan.
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