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Transcendental calc

  1. May 28, 2007 #1
    [​IMG] [Broken]

    I appreciate any help for these 2 problems.

    EDIT: First problem is solved!
     
    Last edited by a moderator: Apr 22, 2017 at 5:51 PM
  2. jcsd
  3. May 28, 2007 #2
    To evaluate
    [tex]\int \frac{1}{e^x + 1} dx[/tex]
    First make the substitution [tex]u=e^x +1[/tex] and work through to obtain
    [tex]\int \frac{1}{u^2 -u} du[/tex]

    Next http://faculty.ed.umuc.edu/~swalsh/Math%20Articles/GeomCS.html [Broken] on the bottom line to obtain.

    [tex]\int \frac{1}{(u-\frac{1}{2})^2 -\frac{1}{4}} du[/tex]

    Now let [tex]y=u-\frac{1}{2}[/tex] to obtain
    [tex]\int \frac{1}{y^2 -(\frac{1}{2})^2} dy[/tex]
    This is a known integral, you should have it in your log tables and eventually with practice will become somewhat familiar with it. This integral is

    [tex]\int \frac{1}{y^2 -a^2} dy=\frac{1}{2a}\log\left(\frac{y-a}{y+a}\right)[/tex]
    So here a=1/2 and the integral is

    [tex]\log\left(\frac{y-\frac{1}{2}}{y+\frac{1}{2}}\right)[/tex]
    Putting back u
    [tex]\log\left(\frac{u-1}{u}\right)[/tex]
    Finally put back x
    [tex]\log\left(\frac{e^x}{e^x +1}\right)[/tex]
    Which you can check is the right answer. You should check the differenciation to get a feel for how the derivative works here.

    The key to this question was knowing how to complete the square for the inverse polynomial part and of course being willing to make a second substitution.
     
    Last edited by a moderator: Apr 22, 2017 at 5:51 PM
  4. May 29, 2007 #3

    Gib Z

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    An easier way would have been to multiply the numerator and denominator by e^(-x) and let u = e^(-x)
     
    Last edited: May 29, 2007
  5. May 29, 2007 #4
    An impressive response that cleared everything up. Thank you so much!
     
    Last edited by a moderator: Apr 22, 2017 at 5:52 PM
  6. May 30, 2007 #5

    Gib Z

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

    [itex]u= e^{-x}, du = -e^{-x} dx[/itex]

    [tex] - \int \frac{1}{1+u} du = -\log_e (1+u) = -\log_e ( 1+e^{-x})[/tex]
     
    Last edited: May 30, 2007
  7. May 30, 2007 #6

    matt grime

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    Why make that subsitution at all, GIb. Before you set e=e^-x, the integrand is of the form f'/f, so it integrates to log |f|.
     
  8. May 30, 2007 #7

    Gib Z

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    Either way, quicker than the 2 substitutions, completing the square and requiring a table of standard integrals.
     
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