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

1. anonymous

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

2. anonymous

u = 1+e^6?

3. anonymous

Good guess, go for it.

4. anonymous

Do you mean 1+e^x?

5. anonymous

yes, thats what I meant, sorry

6. 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.

7. 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.

8. anonymous

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

9. 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.

10. 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)$

11. anonymous

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

12. 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.

13. anonymous

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

14. anonymous

It's in the answer section, no explanation.

15. anonymous

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

16. anonymous

Yeah, it says +

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

21. anonymous

^^

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

26. anonymous

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

27. anonymous

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