## anonymous 5 years ago Integral of 1 / ( 1 + e ^ x) actual process please... :D

1. anonymous

Dang you got 103 fans now!

2. anonymous

Hi arman, Try looking at it like this$\int\limits_{}{}\frac{1}{1+e^x}dx=\int\limits_{}{}\frac{1+e^x-e^x}{1+e^x}dx=\int\limits_{}{}1-\frac{e^x}{1+e^x}dx$

3. anonymous

Haha, I know!

4. anonymous

$=x-\log (1+e^x) + c$

5. anonymous

log == ln always unless a base is made explicit

6. anonymous

It's an arctrig function and x-ln(1+e^x) is e^x/(1+e^x) (chain rule)

7. anonymous

II know that it's arctan of something ><

8. anonymous

The second integral makes use of this fact:$\int\limits_{}{}\frac{f'(x)}{f(x)}dx=\log f(x)+c$

9. anonymous

If f(x)=1+e^x, then f'(x)=e^x, which is the form you have after rearranging.

10. anonymous

I'm on my fone so I didn't see half of those equations till after I posted my last post.... lemmy try that... yummy and sec

11. anonymous

Gimmy a sec**

12. anonymous

u substitution then partial fraction decomposition

13. anonymous

I don't get why I can't think of these on my own -.-

14. anonymous

i got an answer of ln(e^x/(e^x+1))+c

15. anonymous

There's one way to check your answer - take the derivative and see if you get back to the integrand.

16. anonymous

yea it does

17. anonymous

just checked it

18. anonymous

Nope, pfd is in the chapter after this (I slipped it previousy and went ahead) so Lucas answer makes perfect sense...

19. anonymous

Lol

20. anonymous

Good work

21. anonymous

Oh boy, integral of e^sqrtx from 0 to 4...

22. anonymous

one sec

23. anonymous

I don't even know what to use on this one...

24. anonymous

2(e^(sqrt(x))(sqrt(x)-1)|(0,4)

25. anonymous

Why?

26. anonymous

have u learned integration by parts yet?

27. anonymous

Yep

28. anonymous

and i take it you know u substitution?

29. anonymous

30. anonymous

Ok...

31. anonymous

u should get int(2u(e^u)du)

32. anonymous

do you follow me so far?

33. anonymous

You can put the dx in terms of u aswell? (Like I can write 2u instead of 2sqrtx? )

34. anonymous

correct

35. anonymous

from this point you can use integration by parts

36. anonymous

Ok then that makes sense and i can use the uv vdu then.... -.- I'm retarded

37. anonymous

lol its hard to see that without substituting first tho

38. anonymous

No... I'm just retarded. And the answer is 2e^2

39. anonymous

did u evaluate at zero also?

40. anonymous

im getting 2(e^2)+2

41. anonymous

Good point, further testing my retardation.

42. anonymous

lol common mistake

43. anonymous

When you uv vdu sub, can you choose which term is your u and v? I'm pretty sure you can

44. anonymous

usually it depends on what you have

45. anonymous

to remember what to take for u i use "lipet"

46. anonymous

which gives you an order in which to take your u (Lograthims Inverse.trg Polynomials Exponetials Trigonometrics)

47. anonymous

Well I got a badass equation here: integral of (lnx) /x^2 from 1 to 2

48. anonymous

I know I gotta use uv vdu on it

49. anonymous

judging by lipet take log to be your u

50. anonymous

and 1/x^2 to be your dv

51. anonymous

im getting (-ln(x)/x)-(1/x)+c

52. anonymous

your evaluated integral should come out to be (-ln(2)/2)+1/2

53. anonymous

were you able to solve it?