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Integration - 2

  1. May 15, 2004 #1
    [tex]\int_{1}^{2} \frac{x^2 - 2x - 9}{(2x-1)(x^2+3)} dx[/tex]

    i already get theanswer [tex]\searrow -1.09 [/tex]

    then, given that

    y = \frac{x^2 - 2x - 9}{(2x-1)(x^2+3)}

    find \frac{dy}{dx} [/tex] when x=-1

    how to use the answer from previous stage to apply into this question?
  2. jcsd
  3. May 15, 2004 #2
    I think they are different, so if
    [tex]\frac{dy}{dx} = \frac{(\frac{dU}{dx}V)-(\frac{dV}{dx}U)}{V^2}[/tex]

    >>>>> Sorry changed plus to minus. really stupid I am
    Last edited: May 15, 2004
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