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An integral

  1. Dec 14, 2006 #1

    ddr

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    1. The problem statement, all variables and given/known data

    integral defined in 0 (down) and 1 (up) of:
    (16x^4 - 4)/(4x^2+1)

    2. Relevant equations



    3. The attempt at a solution

    maybe the partition mode?
    who help me?
     
  2. jcsd
  3. Dec 14, 2006 #2
    Just wondering...by any chance, is the denominator of the integrand 4x2+2?
     
    Last edited: Dec 14, 2006
  4. Dec 14, 2006 #3

    arildno

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    Dearly Missed

    Hint:
    Use polynomial division first.
     
  5. Dec 15, 2006 #4

    ddr

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    i can use the substitution rule, with u=4x^2?
     
  6. Dec 15, 2006 #5

    HallsofIvy

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    You could but it doesn't really help since du= 8xdx doesn't give you anything easy. The numerator obviously factors into (4x2-2)(4x2+2)- that's why arildno asked if the denominator wasn't actually 4x2+ 2 rather than 4x2+ 1. But the world is never that easy, not even homework problems.

    Best thing to do is arildno's suggest. Go ahead and divide 16x4- 4 by 4x2+ 1. The result will be a cubic polynomial plus a linear term, Ax+ B, over 4x2+ 1. To integrate Ax/(4x2+1), let u= 4x2+ 1. To integrate B/(4x2+1), use the arctangent.
     
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