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Homework Help: Partial fraction decomposition

  1. Jun 18, 2008 #1
    Problem: Find the integral of 2dx/(2x-3) using partial fraction decomposition
    Compare coefficients A = 2
    [tex] 2 \int (1/(2x-3)) = 2ln(2x-3)[/tex]
    This is wrong. Can someone please tell me what I'm doing wrong?
  2. jcsd
  3. Jun 18, 2008 #2
    Use u-substitution. Let u=2x-3.
  4. Jun 18, 2008 #3


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    you don't need partial fractions for this.

    just put u=2x-3 and go from there
  5. Jun 18, 2008 #4
    This is just a simple problem that I want to apply the technique to and further my understanding of it so that I can begin using it on more difficult questions.
  6. Jun 18, 2008 #5
    Did the problem tell you that you must use partial fraction for this problem? Because in my opinion, it makes no sense to use it.

    Also, [tex]\int \frac{1}{2x-3} \neq ln(2x-3)+C[/tex] therefore [tex]2 \int \frac{1}{2x-3} \neq 2ln(2x-3)+C[/tex]
    Last edited: Jun 18, 2008
  7. Jun 18, 2008 #6


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    well in a sense...when you have a linear factor in the denominator like 2x-3, in the question, it kinda already is a partial fraction.
  8. Jun 18, 2008 #7
    The integral of 1/x is lnx + c
    So then shouldnt the integral of
    1/2x-3 be ln(2x-3) + c
    If not then what is it?
    And yes the question did ask me to use partial fractions for this problem.
  9. Jun 18, 2008 #8
    Integrals of the form a+bu

    [tex]\int \frac{1}{a+bu}du=\frac{1}{b}ln|a+bu|+C[/tex]

    To check that this is false: if you take the derivative with respect to x of "ln(2x-3) + c", it will not equal [tex]\frac{1}{2x-3}[/tex] since you have to use the chain rule.
    Last edited: Jun 18, 2008
  10. Jun 18, 2008 #9


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    [tex]\int \frac{1}{2x-3} dx[/tex]

    if u=2x-3, then [itex]\frac{du}{dx}=2 \Rightarrow \frac{du}{2}=dx[/itex]

    now just put that into the integral
  11. Jun 18, 2008 #10
    Can partial fraction decomposition even be used in this situation to get a correct answer?
  12. Jun 18, 2008 #11


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    The partial fraction decomposition of 1/(2x-3) IS 1/(2x-3). Please believe what people are telling you.
  13. Jun 19, 2008 #12
    How would i do this one:
    integral of ((x^2)+1)/(6x-9x^2)
    The highest degree of the denominator is equal to the highest degree of the numerator but it has to be greater than it.
  14. Jun 19, 2008 #13


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    Do polynomial long division first, to make sure the degree of the denominator is greater than that of the denominator.
  15. Jun 19, 2008 #14
    Have no idea how to do that. Can you please show me?
  16. Jun 19, 2008 #15

    Gib Z

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    What, you mean how to divide?

    Well, see if you can follow this:

    [tex] \frac{x^2 +1}{6x-9x^2} = \frac{-1}{9}\left( \frac{9x^2+9}{9x^2 - 6x} \right)

    = \frac{-1}{9}\left( \frac{9x^2 -6x + 6x +9}{9x^2 - 6x} \right)

    = \frac{-1}{9}\left( 1 + \frac{6x+9}{9x^2 - 6x} \right)

    = \frac{-1}{9}\left( 1+ \frac{2x+3}{3x^2 - 2x} \right)[/tex].

    How about now?
  17. Jun 19, 2008 #16
    i understand but why did the sign of only the denominator change in the 1st step and not the numerator?
    Edit: nevermind, figured it out.
    THanks for your help everyone
    Last edited: Jun 19, 2008
  18. Jun 19, 2008 #17
    http://www.imagehuge.com/images/a7ttelgdzf3crvdsouk.jpg [Broken]
    Last edited by a moderator: May 3, 2017
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