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Math Help - partial fractions integral

  1. #1
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    partial fractions integral



    I'm getting a slightly different answer: I don't come up with the x^2, but I've got everything else. Any ideas what I could have missed?

    I went the 2x^3-4x^2-15x+5=A(x-4)+B(x+2) route and got A=\frac{-1}{2} and B=\frac{3}{2}.
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  2. #2
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    The steps the book follows confuse me, too. You can see them at calcchat.com|Textbook: Calculus, 8th Edition|Chapter 8|Section 5|Exercise 15.

    Such as where that 2x and 5+x came from in step 1.
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    Quote Originally Posted by dataspot View Post


    I'm getting a slightly different answer: I don't come up with the x^2, but I've got everything else. Any ideas what I could have missed?

    I went the 2x^3-4x^2-15x+5=A(x-4)+B(x+2) route and got A=\frac{-1}{2} and B=\frac{3}{2}.
    Since the degree of the numerator is larger than the degree of the denominator you have to do a long division and divide the quadratic into the cubic before attempting to get a partial fraction decomposition. The result of the long division is

    2x + \frac{x+5}{x^2 - 2x - 8}.

    Now you can attempt to get a partial fraction decomposition ........ of \frac{x+5}{x^2 - 2x - 8} = \frac{x + 5}{(x - 4)(x + 2)}.
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