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Integral: sqrt(1+((x^4-1)/(2x^2))^2)

  1. Dec 26, 2009 #1
    http://integrals.wolfram.com/index.jsp?expr=sqrt(1%2B((x^4-1)%2F(2x^2))^2)&random=false

    Can someone explain me how to solve an integral like this?
    I tried it, but I have absolutely no idea how to solve it..
     
  2. jcsd
  3. Dec 26, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Alexx1! Welcome to PF! :smile:
    Do you mean [tex]\sqrt(1+(\frac{x^4-1}{2x^2})^2)[/tex] ? :confused:
     
  4. Dec 26, 2009 #3
    Re: Welcome to PF!

    Yes, it's that integral ;)
     
  5. Dec 26, 2009 #4

    tiny-tim

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    ok, then (without integrating) expand everything inside the √, and then simplify …

    what do you get? :smile:
     
  6. Dec 26, 2009 #5
    (8x^8+2x^4+x)/(4x^4)
     
  7. Dec 26, 2009 #6

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)
    uhhh? :confused: not even close :redface:
     
  8. Dec 26, 2009 #7
    ((x^4-1)/2x^2)^2) = ((x^4-1)^2)/((2x^2)^2)

    = (x^8+1-2x^4)/(4x^4)

    ==> 1 + (x^8+1-2x^4)/(4x^4) = (4x^4+x^8+1-2x^4)/(4x^4)

    = (x^8+2x^4+1)/(4x^4)

    This is the correct answer?
    I made a mistake it had to be: x^8 (instead of 8x^8) and 1 (instead of x)
     
  9. Dec 26, 2009 #8

    tiny-tim

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    (please use the X2 tag just above the Reply box … it makes it much easier to read)
    ok! :smile:

    and the square-root of that is … ?​
     
  10. Dec 26, 2009 #9

    Mentallic

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    What should've made factorising easy to spot is that you first expanded
    [tex](x^4-1)^2=x^8-2x^4+1[/tex]

    and after adding [itex]4x^4[/itex] you now have
    [tex]x^8+2x^4+1[/tex]

    Isn't it clear how this can be factored?
     
  11. Dec 26, 2009 #10

    HallsofIvy

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    The crucial point is that [itex]4a+ (x- a)^2= 4a+ x^2- 2ax+ a^2[/itex][itex]= x^2+ 2ax+ a^2= (x+a)^2[/itex].

    That's used a lot to produce easy "arc length" problems!
     
  12. Dec 26, 2009 #11
    = √x8+2x4+1/√4x4

    = √(x4+1)2/2x2

    = x4+1/2x2
     
  13. Dec 26, 2009 #12

    tiny-tim

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    come on, think! :smile:

    (or put x4 = y :wink:)
     
  14. Dec 26, 2009 #13
    Ok, I found it!

    = √x8+2x4+1/√4x4

    = √(x4+1)2/2x2

    = x4+1/2x2

    ==> Integral x4+1/2x2

    = (1/2) Integral (x4/x2) + (1/2) Integral (1/x2)

    And so on..
     
  15. Dec 26, 2009 #14
    Thanks everyone for helping me out!
     
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