1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    Let:

    [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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook