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Indefinite Integration by exchange of variables

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \int \frac {2+z^{-1}}{z^{2}} dz [/tex]

    3. The attempt at a solution


    [tex] u = 2 +z^{-1} [/tex]

    [tex] du = -z^{-2} dz [/tex]

    [tex] dz = -z^{2} du [/tex]

    so now its

    [tex] \int \frac {u}{z^{2}} (-z^{2}) du [/tex]

    [tex] \int \frac {(u)(-z^{2})}{z^{2}} du [/tex]

    [tex] \int (u)(-1) du [/tex]

    and then the antiderivative of u*(-1) is

    [tex] -\frac{1}{2}(2+z^{-1})^{2} + C [/tex]

    right? The answer in the book is:

    [tex] -2z^{-1}-\frac{1}{2}z^{-2} + C [/tex]

    I don't see anywhere that I went wrong....
  2. jcsd
  3. Apr 28, 2010 #2
    expand your parenthesis and put the -2 that results in with the constant. You have the same answer, just in a different form.
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