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Homework Help: Length of a curve

  1. Jan 24, 2008 #1
    1. The problem statement, all variables and given/known data

    I need to find the length of [tex]\frac{\frac{1}{3}x^{3} + x^{2} + x + 1}{4x+4}[/tex] from x=0 to x=2 but i can not factor this down to be able to set up the integral, any suggestions, here is the derivative:
    [tex]\frac{\frac{8}{3}x^{3}+8x^{2}+8x}{16x^{2}+32x+16}[/tex]
    It is possible I might have to factor the derivative rather than the function itself which is why i supplied both.
     
  2. jcsd
  3. Jan 25, 2008 #2

    Defennder

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  4. Jan 25, 2008 #3

    HallsofIvy

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    I would write this as:
    [tex]\frac{1}{12} \frac{(x^3+ 3x^2+ 3x+ 1)+ 2}{x+1}=\frac{1}{12}\frac{(x+1)^3+ 2}{x+1}= \frac{1}{12}((x+1)^2+ \frac{2}{x+1})[/tex]
    That looks to me like it will be easier to handle.
     
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