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Integral involving square root and exp

  1. Mar 2, 2014 #1
    1. The problem statement, all variables and given/known data

    [itex]\int[/itex][itex]\frac{dx}{\sqrt{e^{x} + 1}}[/itex]

    2. Relevant equations
    Using u-substitution


    3. The attempt at a solution

    Let u = [itex]\sqrt{e^{x} + 1}[/itex] [itex]\Rightarrow[/itex] u[itex]^{2} - 1[/itex] = e[itex]^{x}[/itex]
    Then, du = [itex]\frac{e^{x} dx}{2\sqrt{e^{x} + 1}}[/itex] [itex]\Rightarrow[/itex] dx = [itex]\frac{2u du}{u^{2}-1}[/itex]

    So, [itex]\int[/itex][itex]\frac{dx}{\sqrt{e^{x} + 1}}[/itex] = [itex]\int[/itex][itex]\frac{2u du}{u(u^{2}-1)}[/itex]

    But, I'm stuck at this point. I think I want to break it up into two simpler integrals, but I'm not sure how to do this. Any suggestions would be greatly appreciated!
     
  2. jcsd
  3. Mar 2, 2014 #2

    Dick

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    Homework Helper

    Partial fractions is what you want. Factor the denominator.
     
  4. Mar 2, 2014 #3
    Okay.

    So, [itex]\int[/itex][itex]\frac{dx}{\sqrt{e^{x} + 1}}[/itex] = [itex]\int[/itex][itex]\frac{2u du}{u(u^{2}-1)}[/itex] = [itex]\int[/itex][itex]\frac{2du}{(u+1)(u-1)}[/itex]

    And now I'm stuck again.
     
  5. Mar 2, 2014 #4

    Dick

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    Partial fractions! ##\frac{1}{(u+1)(u-1)}=\frac{A}{u+1}+\frac{B}{u-1}## for some constants A and B. Find those constants.
     
  6. Mar 2, 2014 #5
    Okay. I think I've got it.

    So, [itex]\int[/itex][itex]\frac{dx}{\sqrt{e^{x} + 1}}[/itex] = [itex]\int[/itex][itex]\frac{2u du}{u(u^{2}-1)}[/itex] = [itex]\int[/itex][itex]\frac{2du}{(u+1)(u-1)}[/itex] = [itex]\int[/itex][itex]\frac{du}{u-1}[/itex] - [itex]\int[/itex][itex]\frac{du}{u-1}[/itex] = ln|u-1| - ln|u+1| = ln[itex]\frac{|u-1|}{|u+1|}[/itex] where u = [itex]\sqrt{e^{x}+1}[/itex]

    = ln[itex]\frac{\sqrt{e^{x}+1}-1}{\sqrt{e^{x}+1}+1}[/itex]

    Thank you for all your help.
     
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