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Integration with partial fractions - help!

  1. Nov 11, 2009 #1
    Integration with partial fractions -- help!

    1. The problem statement, all variables and given/known data
    Here is the problem: http://img130.imageshack.us/img130/1673/integralthing.png [Broken]
    The answer should be 4.


    2. Relevant equations

    N/A

    3. The attempt at a solution

    Here are my steps so far:

    (5x^2 - 17x + 10) / ((x-1)^3 * (x+1)) = A/(x-1) + B/(x-1)^2 + C/(x-1)^3 + D/(x+1)
    5x^2 - 17x + 10 = A(x-1)^2(x+1) + B(x-1)(x+1) + C(x+1) + D(x-1)^3

    Substituting 1 for x gives:
    5-17+10 = 2C
    C = -1

    Substituting -1 for x gives:
    5(-1)^2 - 17(-1) + 10 = -8D
    D = -4

    Variables C and D are correct according to the answer given to me on the practice test.

    Now I'm at the point where I want to find the other 2 variables, A and B. I'm not sure how to go about doing this. I talked to my professor and she said something about plugging in random values for x and somehow getting A and B that way, but it didn't make much sense to me after she explained it. I have to be able to do this for a test where I will not be allowed anything but the Apple OSX calculator (it's basically the same as the Windows Calculator), so the easiest method to do this would be appreciated.

    Thank you,
    Bensky
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 11, 2009 #2

    Mark44

    Staff: Mentor

    Re: Integration with partial fractions -- help!

    You're on the right track with the way you split up the separate fractions. The important thing is that your equation -- (5x^2 - 17x + 10) / ((x-1)^3 * (x+1)) = A/(x-1) + B/(x-1)^2 + C/(x-1)^3 + D/(x+1) -- has to hold for all values of x. That's what your teacher was talking about when he/she said to plug in random numbers. Presumably you have already tried x = 1 and x = -1. Another one to try is x = 0. Your choice for another one.
     
  4. Nov 11, 2009 #3
    Re: Integration with partial fractions -- help!

    Thanks Mark -- I actually figured this out on my own a minute or so before I saw this post.

    For anyone else trying to do these types of problems: I ended up substituting x=2 and x=-2 to get the system of eq's (solved by elimination):
    -3A + B = -15
    3A + 3B = 3
    -------------------
    The A's already cancel, so...
    4B = -12
    B=-3 => A = 4

    Maybe I should have screwed around a bit more before I posted here. :blushing:
     
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