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Homework Help: Integration of rational functions

  1. Mar 6, 2004 #1
    How do I solve the integration of a rational function such as:

    x^2 - 6x - 2
    (x^2 + 2)^2

    If possible, please list the general rule of solving, I DO NOT want the answer, I simply want to know the way of solving it.
    Thanks in advance!
  2. jcsd
  3. Mar 7, 2004 #2


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    Science Advisor

    The "rule" that you want is "partial fractions".

    If you have a rational function, in which the denominator can be factored into distinct linear factors, such as
    (x- 3)/((x+1)(x-2)), then you can write it as a sum of fractions, each having one factor as denominator:
    (x-3)/((x+1)(x-2))= A/(x+1)+ B/(x-2).
    (Of course, the numerator is of lower degree than the denominator: if not, divide first.)

    If some of the linear factors are repeated, such as
    (x+ 4)/((x+1)2(x-2)), then you will need all powers of that repeated factor: A/(x+1)+ B/(x+1)2+ C/(x-2)

    If some of the factors are quadratics that cannot be factored, then they can, by completing the square, be written in the form "a(x-b)2+ c" and you will need a fraction of the form (Ax+ B)/(a(x-b)2+c), for example (3x2- 2x+ 4)/((x2+ 4)(x+3)) can be written (Ax+B)/(x2+4)+ C/(x+3).

    In this particular example,
    [tex]\frac{x^2-6x- 2}{(x^2+2)^2)^2}[/tex]
    can be written in the form

    Those have to be equal for all x so one way of finding A, B, C, D is by setting those equal:
    [tex]\frac{x^2-6x- 2}{(x^2+2)^2)^2}= \frac{Ax+B}{(x^2+2)^2}+\frac{Cx+D}{x^2+2}[/tex]
    Now multiply both sides by that denominator to clear the fractions and set x equal to 4 different numbers to get 4 equations for A, B, C, and D. You can often choose those numbers to simplify the equations.
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