Indefinite integral

  • Thread starter cmab
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  • #1
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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???? !
 

Answers and Replies

  • #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
 
  • #3
32
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p53ud0 dr34m5 said:
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

I dont get it :confused:
 
  • #4
2,209
1
[tex] \frac{1}{e^x + e^{-x}} = \frac{1}{\frac{(e^x)^2+1}{e^x}}[/tex]

Can you go from there?
 
  • #5
James R
Science Advisor
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This might help you:

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

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