anonymous
  • anonymous
How to integrate 1/(1+e^x) from (0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Try u substitution. What would you guess? What would you let u be?
anonymous
  • anonymous
u = 1+e^6?
anonymous
  • anonymous
Good guess, go for it.

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anonymous
  • anonymous
Do you mean 1+e^x?
anonymous
  • anonymous
yes, thats what I meant, sorry
anonymous
  • anonymous
Aaronip, that actually won't work. This problem is actually pretty difficult. I can't think of anything to say but try u = e^x because that will work. However, you have to figure out how to use u in that way. Then you need to use partial fractions.
anonymous
  • anonymous
Yeah, this a special case.Let u be x. so integrate du/(1+e^u). It turns out this is equal to (-e^-u du)/((e^-u)+1). I don't know why. I have to research this, getting answer out of a book. Integrated it becomes ln |1+e^-u|. Evaluate over interval; there is a change of interval with the x to u thing. Good luck.
anonymous
  • anonymous
Okay, I'll try that. Thank you so much~!
anonymous
  • anonymous
Actually, it you can do it as u = e^x and then integrate du/(u*(1+u)) and then do partial fractions.But there might be an easier way.
anonymous
  • anonymous
aaronip, there's an easier way. Notice that\[\frac{1}{1+e^x}=\frac{1+e^x-e^x}{1+e^x}=\frac{1+e^x}{1+e^x}-\frac{e^x}{1+e^x}=1-\frac{e^x}{1+e^x}\]Then you'd have\[\int\limits_{0}^{1}\frac{dx}{1+e^x}=\int\limits_{0}^{1}1-\frac{e^x}{1+e^x}dx=\left[ x-\log (1+e^x) \right]_0^1\]\[=(1-\log (1+e))-(0-\log 2)=1+\log 2-\log (1+e)\]
anonymous
  • anonymous
Lokisan~ You're amazing~! Thank you~!
anonymous
  • anonymous
Good job lokisan I would like to see the machinations that make 1/(1+e^x) equal to (-e^-u du)/((e^-u)+1) as I found in a book.
anonymous
  • anonymous
haha, thank you :) Umm, what's your book thinking?
anonymous
  • anonymous
It's in the answer section, no explanation.
anonymous
  • anonymous
Are you sure the + sign in the denominator should be there?
anonymous
  • anonymous
Yeah, it says +
anonymous
  • anonymous
I get\[-\frac{e^{-u}}{e^{-u}-1}\] if I make the substitution,\[e^u = 1+e^x\]
anonymous
  • anonymous
I'm looking at it. to see what you did. One minute
anonymous
  • anonymous
I don't get it. what did you do?
anonymous
  • anonymous
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anonymous
  • anonymous
^^
anonymous
  • anonymous
Got it. So may the book distributed the negative sign at the bottom?
anonymous
  • anonymous
Maybe...that's assuming my guess for the book's substitution is correct.
anonymous
  • anonymous
Good job brilliant. Where did you learn these little tricks?
anonymous
  • anonymous
advanced degree in mathematics
anonymous
  • anonymous
Hoo, hoo. One day I will be up there.
anonymous
  • anonymous
Good! I'm off. Enjoy your book ;p

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