1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    [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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Partial fraction decomposition