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Integration (Partial Fractions)

  1. Jan 4, 2006 #1
    [tex]\int \frac{x^2 + 2x}{x^3 + 3x^2 + 4} dx [/tex]

    I can solve it directly by using substitution . But how to solve it by using partial fraction? Is it possible?
     
  2. jcsd
  3. Jan 5, 2006 #2

    Tide

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    You can evaluate the integral using partial fractions if you can factor the polynomial in the denominator.
     
  4. Jan 5, 2006 #3
    I cant factor out the denominator of this function, that means this question cant be solved by using partial fraction?
     
  5. Jan 5, 2006 #4

    TD

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    Not necessarily, you could still factor it but then it would involve radicals and that's usually not very 'neat'. I think you should just praise yourself lucky that the substitution works well here, with the nominator being the denominator's derivative (up to a factor 3) :smile:
     
  6. Jan 5, 2006 #5
    I don't see any reason to use partial fractions when this looks like a cookie cutter substitution problem.
     
  7. Jan 5, 2006 #6
    I know the problem can be solved by substitution. I am just asking whether it can be solved by using partial fraction or not.
     
  8. Jan 5, 2006 #7

    Tx

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    I don't see why this would need partial fractions, the bottom is one degree higher than the top. That hints towards log laws...
     
  9. Jan 6, 2006 #8

    VietDao29

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    ???
    Are you sure?
    This integral:
    [tex]\int \frac{x ^ 2}{x ^ 3 + x ^ 2 - x - 1} dx[/tex]
    also has the denominator that is 1 degree higher than the numerator, can it be solve by log laws, without using partial-fraction???
     
  10. Jan 7, 2006 #9

    Tx

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    Touche.
    That one is a classic instance of partial fractions. However, this one is a log law one cause the number of terms is in preportion with the original number.
     
  11. Jan 7, 2006 #10

    shmoe

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    What do you mean by this?

    More is needed than just degrees or the number of terms being 'correct' for it to work out as a log. The original question is in the form K*P'(x)/P(x), where K is a constant, so it's a logarithmic derivative.

    The denominator of the original question can be factored as TD says, look up the cubic formula. It's gross though, and worth looking at to see how 'lucky' you were that this question works out in a nicer way.
     
  12. Jan 7, 2006 #11

    Tx

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    You are right, absolutely. This question should use partial fractions, however, The derivative was on the top so I just said that it looks like a log integral. I apologise if what I was saying was utter garbage.

    What I meant was that the top was very similar to the derivative of the bottom. I just worded it incorrectly.
     
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