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anonymous
 5 years ago
How to integrate 1/(1+e^x) from (0<x<1)
anonymous
 5 years ago
How to integrate 1/(1+e^x) from (0<x<1)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Try u substitution. What would you guess? What would you let u be?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good guess, go for it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, thats what I meant, sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Aaronip, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I'll try that. Thank you so much~!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, 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
 5 years ago
Best ResponseYou've already chosen the best response.0aaronip, there's an easier way. Notice that\[\frac{1}{1+e^x}=\frac{1+e^xe^x}{1+e^x}=\frac{1+e^x}{1+e^x}\frac{e^x}{1+e^x}=1\frac{e^x}{1+e^x}\]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
 5 years ago
Best ResponseYou've already chosen the best response.0Lokisan~ You're amazing~! Thank you~!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good 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
 5 years ago
Best ResponseYou've already chosen the best response.0haha, thank you :) Umm, what's your book thinking?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's in the answer section, no explanation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you sure the + sign in the denominator should be there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I get\[\frac{e^{u}}{e^{u}1}\] if I make the substitution,\[e^u = 1+e^x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm looking at it. to see what you did. One minute

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't get it. what did you do?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Got it. So may the book distributed the negative sign at the bottom?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Maybe...that's assuming my guess for the book's substitution is correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good job brilliant. Where did you learn these little tricks?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0advanced degree in mathematics

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hoo, hoo. One day I will be up there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good! I'm off. Enjoy your book ;p
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