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Calculus Question

  1. May 28, 2005 #1

    VietDao29

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    Hi,
    Can you guys just throw me a hint on how to integrate:
    [tex]\int \frac{x^2 - 1}{x^4 + 1}dx[/tex]
    I am completely lost. I don't even have any idea how to start. :confused:
    Any help will be appreciated.
    Thanks.
    Viet Dao,
     
  2. jcsd
  3. May 28, 2005 #2
    Hmmm... stupid plus sign!

    I did this on an online integrator and got the difference of two logs, which would make me think partial fractions, although in this case that would deal with complex numbers.
     
  4. May 28, 2005 #3

    Pyrrhus

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    You can solve this with Partial fractions.

    [tex] x^4 + 1 = (x^2 - \sqrt{2}x +1) (x^2 + \sqrt{2}x +1) [/tex]
     
  5. May 28, 2005 #4
    That's it. The answer online used that same form.
     
  6. May 28, 2005 #5

    HallsofIvy

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    To see how Cyclovenom got that, if you let y= x2, x4+ 1 becomes y2+ 1. The roots to y2+ 1= 0 are y= +/- i. That means we have y= x2= i so that [tex]x= \frac{\sqrt{2}}{2}(1+i)[/tex] and [tex]x= -\frac{\sqrt{2}}{2}(1+i)[/tex]. We also have y= x2= -i so that
    [tex]x= \frac{\sqrt{2}}{2}(1- i)[/tex] and [tex]x= -\frac{\sqrt{2}}{2}(1- i)[/tex], the four complex roots of x4+ 1= 0. That tells us that x4+ 1 factors as [tex](x-\sqrt{2}}{2}(1+i))(x+\sqrt{2}}{2}(1+i))(x-\frac{\sqrt{2}}{2}(1- i))(x+\frac{\sqrt{2}}{2}(1- i))[/tex].
    We can rearrange the factors as [tex](x-\sqrt{2}}{2}(1+i))(x-\frac{\sqrt{2}}{2}(1- i))(x+\sqrt{2}}{2}(1+i))))(x+\frac{\sqrt{2}}{2}(1- i))[/tex] to get the two real factors Cyclovenom gives.
     
  7. May 28, 2005 #6
    Or you can just say that

    x^4 + 1 = (x^2 +/- ax +/- 1)(x^2 -/+ ax +/- 1)

    and in a few tries you get that a = sqrt(2).

    Trial and error isn't elegant, but in this case it's a lot faster.
     
  8. May 28, 2005 #7

    dextercioby

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    How about that

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

    and then use the square difference formula...?

    Daniel.
     
  9. May 28, 2005 #8
    Daniel,

    Very elegant and very fast. As usual, you win!

    jdl
     
  10. May 28, 2005 #9

    GCT

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    I'm not quite sure what you did here. Can you or anyone else here point out where I might learn the details of this? References, such as texts, online sites, the theorems, title of the subject, or even a brief explanation.
     
  11. May 29, 2005 #10

    VietDao29

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    I get it now. It's not as 'hard' as I used to think. :smile:
    Thanks very much,
    Viet Dao,
     
  12. May 29, 2005 #11
    could you use u-substition to find the answer too?
     
  13. May 29, 2005 #12
    Nope, at least it doesn't look like it.
     
  14. May 29, 2005 #13
    If you take the next two numbers in this sequence and multiply them together, what number will you get?

    594, 487, 566, 493, 310, 447, ____, ____
     
  15. May 30, 2005 #14
    Good try. You've already made a threadt for this topic. Don't double post. It makes people mad here and it's against the rules.
     
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