Question

How to integrate 1/(1+e^x) from (0 15 years ago

Answer 1

Try u substitution. What would you guess? What would you let u be?

Answer 2

u = 1+e^6?

Answer 3

Good guess, go for it.

Answer 4

Do you mean 1+e^x?

Answer 5

yes, thats what I meant, sorry

Answer 6

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.

Answer 7

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.

Answer 8

Okay, I'll try that. Thank you so much~!

Answer 9

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.

Answer 10

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\[\int\limits_{0}^{1}\frac{dx}{1+e^x}=\int\limits_{0}^{1}1-\frac{e^x}{1+e^x}dx=\left[ x-\log (1+e^x) \right]_0^1\]\[=(1-\log (1+e))-(0-\log 2)=1+\log 2-\log (1+e)\]

Answer 11

Lokisan~ You're amazing~! Thank you~!

Answer 12

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.

Answer 13

haha, thank you :) Umm, what's your book thinking?

Answer 14

It's in the answer section, no explanation.

Answer 15

Are you sure the + sign in the denominator should be there?

Answer 16

Yeah, it says +

Answer 17

I get\[-\frac{e^{-u}}{e^{-u}-1}\] if I make the substitution,\[e^u = 1+e^x\]

Answer 18

I'm looking at it. to see what you did. One minute

Answer 19

I don't get it. what did you do?

Answer 20

^^

Answer 21

Got it. So may the book distributed the negative sign at the bottom?

Answer 22

Maybe...that's assuming my guess for the book's substitution is correct.

Answer 23

Good job brilliant. Where did you learn these little tricks?

Answer 24

advanced degree in mathematics

Answer 25

Hoo, hoo. One day I will be up there.

Answer 26

Good! I'm off. Enjoy your book ;p