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Indefinite integral

  1. May 9, 2005 #1
    How do I find the indefinite integral of 1/[(e^x)+(e^-x)] ???

    Do I have to multiply by [(e^x)-(e^-x)]/[(e^x)-(e^-x)] to eliminate the denominator???? !
     
  2. jcsd
  3. May 9, 2005 #2
    are you trying to do:
    [tex]\int\frac{dx}{e^x+e^{-x}}[/tex]

    if so, for a=b=c:
    [tex]\int\frac{dx}{e^{ax}+e^{-bx}}=\int\frac{dx}{e^{cx}+e^{-cx}}=\frac{tan^{-1}(e^{cx})}{c}+C[/tex]
    i hope that helps
     
  4. May 9, 2005 #3

    I dont get it :confused:
     
  5. May 9, 2005 #4
    [tex] \frac{1}{e^x + e^{-x}} = \frac{1}{\frac{(e^x)^2+1}{e^x}}[/tex]

    Can you go from there?
     
  6. May 9, 2005 #5

    James R

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    This might help you:

    [tex]\frac{e^x + e^{-x}}{2} = \cosh x[/tex]
     
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