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March 27, 2017

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Integral of 10 / (5x^2 - 2x^3) Do not determine the numerical values of the coefficients.

Just wanted to know why when I write out the form of the partial fraction decomposition

there is A/x + B/x^3 + C/ (x^2 + 4)

Why isn't it just A/x^3 + B/ (x^2 + 4)?

  • calculus - ,

    your denominator is
    5x^2 - 2x^3
    = x^2(5-2x)

    so you could have had 3 fractions,
    A/x + B/x^2 + C/(5-2x)

    that would have given you a LCD of x^2(5-2x)
    or 5x^2 - 2x^3

    I don't see how you got your choice or even the supposedly correct choice.

  • calculus - ,

    Sorry I typed out the wrong answer. Why would I have 3 fractions though? Because it's x^2?

  • calculus - ,

    yes, the option I gave you was
    A/x + B/x^2 + C/(5-2x)

    of course it could have been only 2 fractions with
    B/x^2 + C/(5-2x) , and the LCD would also have been
    x^2(5-2x)
    but we don't know that, so we have to take the worst-case scenario.

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