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Partial fractions, chain rule

  1. Jun 20, 2005 #1
    Hi, I have 2 questions:

    1. partial fractions:
    if I have following integral: Itegral[(1-2x^2)/(x - x^3)]dx;
    my question is do I break down the denominator to x(1-x^2) or do I go further:
    x(1-x)(1+x); this way it becomes more complicated;

    2. chain rule:
    how does chain rule work in this case: f(x)g(x)h(x)?
    is it something like f'(x)[g(x)h(x)] + g'(x)[f(x)h(x)] + h'(x)[f(x)g(x)] ??? or how would I differentiate this?

    Thanks in advance.
     
  2. jcsd
  3. Jun 20, 2005 #2

    StatusX

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    1. You need enough unknowns to be able to equate the three coefficients for each power of x in the numerator (1,0,-2). So you'll need to break it up into all three factors. In some cases (not this one) the two factors alone would work, but only if the numerator happens to be just right. It would be like finding the point where two lines intersect in 3D; you can only find it if the lines happen to intersect.

    2. Think of it as (fgh)' = f'(gh) + f(gh)'. Your guess is right.
     
  4. Jun 20, 2005 #3
    In first 1-2x^2=1-x^2-x^2 is very helpful. Then you have 2 very easy integrals
     
  5. Jun 20, 2005 #4
    Aaaah, i see. Thanks for help.
     
  6. Jun 20, 2005 #5

    OlderDan

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    #2 is not the chain rule. That is the product rule. It does work as you have it
     
  7. Jun 20, 2005 #6
    Right, sorry, mixed up the names :redface:
    Chain rule is differentiating what I call "embedded functions".
     
  8. Jun 20, 2005 #7
    Composite functions is their real name, btw.
     
  9. Jun 20, 2005 #8
    *mental note to self* TERMINOLOGY!
    I knew it, but did not recall on the spot, thanks everyone!
     
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