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Integral of x/(x^4+1) with respect to t

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data
    integral of x/(x^4+1) with respect to t


    2. Relevant equations
    i know your spose to use (x^2)^2=x^4, and use chain rule
    but apart from that, i have ABSOLUTELY no idea.


    3. The attempt at a solution
    i have no idea where to begin, i tried doing it my reverse product rule but that was an epic fail. if i did i woudlnt be asking for help here.
     
  2. jcsd
  3. Sep 6, 2009 #2

    HallsofIvy

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    Yes, write this as [itex]x/((x^2)^2+ 1)[/itex] and use the substitution [itex]u= x^2[/itex]. Then du= 2x dx of (1/2)du= xdx. The denominator becomes u^2+ 1 so you are now integrating (1/2) du/(u^2+ 1) which should be an easy integral.
     
  4. Sep 6, 2009 #3

    tiny-tim

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    Hi brandy! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    If u = x2, then du = … ? :smile:
     
  5. Sep 6, 2009 #4
    keep in mind im really REALLY slow.
    is this right, you said du= 2x *dx
    so du/2=x*dx
    and u subbed that in
    for the numerator?
    if so, where did the dx go
     
  6. Sep 6, 2009 #5

    tiny-tim

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    ∫ x dx / (x4 + 1)

    = ∫ (1/2) du / (x4 + 1)

    = … ? :smile:
     
  7. Sep 6, 2009 #6
    ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo i get it now
     
  8. Sep 6, 2009 #7

    tiny-tim

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    :biggrin: Wooooooohooooooooooooooooo! :biggrin:
     
  9. Sep 6, 2009 #8
    I might be missing something but doesnt

    integral of x/(x^4+1) with respect to t = [tex]\int \frac{x}{x^4+1} dt[/tex]
     
  10. Sep 6, 2009 #9

    Mark44

    Staff: Mentor

    Or maybe Brandy really means something like this:
    [tex]\int_0^t \frac{x dx}{x^4 + 1}[/tex]

    If that's the case, then this would better be described, in words, as the definite integral from 0 to t of x divided by x4 + 1.

    If that's not the case, then I have no idea what the problem is.
     
  11. Sep 6, 2009 #10
    lol my bad. i was meant to say with respect to x not t
     
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