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Math Help - another partial fractions problem

  1. #1
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    another partial fractions problem

    \int \frac{6x}{x^3-8}dx

    \frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}

    Did I setup this up right? And I'm still not sure how to solve for the variables in this problem...
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  2. #2
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    Yes that is the right setup. You need to fully understand the previous question on the same topic you asked before moving on. You haven't replied to the latest posts there. If you follow the work done in that thread you can see the method that you're asking about. If you still don't get it, ask a question and someone will explain. Then you should be able to do all of these types of problems.
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  3. #3
    Member Mentia's Avatar
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    One tried and true method, once you have your partial fraction set up with your A B C etc is to simply plug in different values for x. If you have 3 variables you need to solve for, A B and C, then try x = 0, 1, 2. You can use whatever values of x you want as long as they are different and don't cause your denominators to go to zero. So in this problem here you would get a system of 3 equations:

    x = 0 -> 0 = -A/2 + C/4
    x = 1 -> -6/7 = -A + B/7 + C/7
    x = 2 -> there is a problem, x = 2 causes a zero in a denominator, try 3 instead
    x = 3 -> 18/19 = A + 3B/19 + C/19

    Now you have a system of 3 equations with 3 unknowns, whats more is they are linear equations so your calculator can solve them ez with a matrix.

    I get A = 1, B = -1, C = 2
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  4. #4
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    A slightly simpler method is to multiply through by the denominators to eliminate the fractions first.
    \frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}= \frac{6x}{x^3- 8} so
    A(x^2+ 2x+ 4}+ B(x)(x-2)+ C(x-2)= 6x
    Setting x= 2, A(4+ 4+ 4)= 12 so A= 1.

    Setting x= 0, 4A+ C(-2)= 0 or -2C= -4 so C= 2.

    Setting x= 1, 7A- B- C= 6 so 7- B- 2= 6 or B= -1.
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