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Integration of rational functions by partial fractions

  1. Sep 10, 2009 #1
    The problem is:

    ((x^3)+x)/(x-1)


    And i need to break it into partial fractions....

    I tried long division and got:

    ((x^2) +x )

    But the book gives me the answer of:

    (x^2)+x+2+(2/(x-1))

    Any help would be very much appreciated, thanks.
     
  2. jcsd
  3. Sep 10, 2009 #2

    arildno

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    Well, IF your long division was right, THEN
    x^3+x=(x^2+x)*(x-1)

    Is that true?
     
  4. Sep 10, 2009 #3

    arildno

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    If you are uncertain of the procedure of long division, try to do it as follows:

    Since the numerator has highest order 3, and the denominator 1, the highest power of the ratio must be..3-1=2.

    We therefore set:

    x^3+x=(Ax^2+Bx+C+f(x))*(x-1), where f(x) is some fractional remainder, and A, B, C are constants to be determined.

    Multiplying out the right-hand side, and ordering in powers of x, we get:

    x^3+x=Ax^3+(B-A)x^2+(C-B)x-C+f(x)*(x-1)

    Thus, we must have A=1.

    B is therefore also 1, in order to cancel the x^2 term.

    C has to be 2, in order for C-B=1

    But then, we must have the last equation:
    0=-C+f(x)*(x-1),
    whereby the remainder f(x)=2/(x-1)

    In total then, we get x^2+x+2+2/(x-1) as our answer.
     
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