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: Integral sign error

  1. Mar 18, 2006 #1

    G01

    User Avatar
    Homework Helper
    Gold Member

    [tex] \int \frac{1+e^x}{1-e^x}dx = [/tex]

    [tex] \int \frac{dx}{1-e^x} + \int \frac{e^x}{1-e^x}dx [/tex]

    The second integral can be done by the substitution
    [tex] u = 1-e^x [/tex]
    [tex] du = -e^x dx[/tex]

    So the second integral becomes:
    [tex]\int \frac{du}{u} = \ln|u|+C[/tex]
    In the first integral, you can use the substituion:
    [tex]w = e^x[/tex]
    [tex]dw = e^x dx[/tex]

    So the first integral becomes:
    [tex] \int \frac{dw}{w(1-w)} [/tex]
    This can be done by parts and you get:
    [tex] \int \frac{dw}{w} + \int \frac{dw}{1-w} [/tex]

    These are also both natural logs so you end up with:

    [tex] \ln|e^x| - 2\ln|1-e^x| + C [/tex]

    [tex] x - 2\ln|1-e^x| +C [/tex]

    I had http://integrals.wolfram.com/index.jsp compute this for me and it got:

    [tex] x - 2\ln(e^x - 1) +C [/tex]

    I think this may have something to do with the absolute value, but I always make stupid mistakes with signs so I thought I'd check.....Thanks for the time I probably just wasted.
     
  2. jcsd
  3. Mar 18, 2006 #2

    G01

    User Avatar
    Homework Helper
    Gold Member

    For some reason it won't let me edit my post..... on the fifth line that integral should be
    [tex]-\int\frac{du}{u} [/tex]
     
  4. Mar 18, 2006 #3

    benorin

    User Avatar
    Homework Helper

    sure, |a-b|=|b-a|
     
  5. Mar 18, 2006 #4

    G01

    User Avatar
    Homework Helper
    Gold Member

    ok thanks alot, stupid question
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook