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anonymous

  • 5 years ago

How to integrate 1/(1+e^x) from (0<x<1)

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  1. anonymous
    • 5 years ago
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    Try u substitution. What would you guess? What would you let u be?

  2. anonymous
    • 5 years ago
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    u = 1+e^6?

  3. anonymous
    • 5 years ago
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    Good guess, go for it.

  4. anonymous
    • 5 years ago
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    Do you mean 1+e^x?

  5. anonymous
    • 5 years ago
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    yes, thats what I meant, sorry

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    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.

  8. anonymous
    • 5 years ago
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    Okay, I'll try that. Thank you so much~!

  9. anonymous
    • 5 years ago
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    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.

  10. anonymous
    • 5 years ago
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    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\[\int\limits_{0}^{1}\frac{dx}{1+e^x}=\int\limits_{0}^{1}1-\frac{e^x}{1+e^x}dx=\left[ x-\log (1+e^x) \right]_0^1\]\[=(1-\log (1+e))-(0-\log 2)=1+\log 2-\log (1+e)\]

  11. anonymous
    • 5 years ago
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    Lokisan~ You're amazing~! Thank you~!

  12. anonymous
    • 5 years ago
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    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.

  13. anonymous
    • 5 years ago
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    haha, thank you :) Umm, what's your book thinking?

  14. anonymous
    • 5 years ago
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    It's in the answer section, no explanation.

  15. anonymous
    • 5 years ago
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    Are you sure the + sign in the denominator should be there?

  16. anonymous
    • 5 years ago
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    Yeah, it says +

  17. anonymous
    • 5 years ago
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    I get\[-\frac{e^{-u}}{e^{-u}-1}\] if I make the substitution,\[e^u = 1+e^x\]

  18. anonymous
    • 5 years ago
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    I'm looking at it. to see what you did. One minute

  19. anonymous
    • 5 years ago
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    I don't get it. what did you do?

  20. anonymous
    • 5 years ago
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  21. anonymous
    • 5 years ago
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    ^^

  22. anonymous
    • 5 years ago
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    Got it. So may the book distributed the negative sign at the bottom?

  23. anonymous
    • 5 years ago
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    Maybe...that's assuming my guess for the book's substitution is correct.

  24. anonymous
    • 5 years ago
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    Good job brilliant. Where did you learn these little tricks?

  25. anonymous
    • 5 years ago
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    advanced degree in mathematics

  26. anonymous
    • 5 years ago
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    Hoo, hoo. One day I will be up there.

  27. anonymous
    • 5 years ago
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    Good! I'm off. Enjoy your book ;p

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