## jotopia34 2 years ago Can anyone help me with the Integral of e^x/(1+e^x) dx

1. Goten77

|dw:1359688972018:dw|

2. Goten77

hmm

3. jotopia34

$\int\limits_{0}^{1}e^x/e ^{2x}+1$

4. jotopia34

sorry I made a mistake its e^2x

5. Goten77

oh then

6. wio

Try u sub

7. wio

$$u=e^x+1$$

8. jotopia34

but it I do u=e^2x, du= 2e^2x and I still can't get rid of the e^x

9. wio

Okay, what exactly is the equation?

10. Goten77

|dw:1359689045480:dw|

11. jotopia34

nope. I still dont have it.....

12. wio

What is the equation?

13. wio

I do not know what the correct equation is.

14. jotopia34

the euation is above

15. wio

There are MANY equations above.

16. wio

Which one is correct?

17. jotopia34

$\int\limits_{0}^{1}e^x/e ^{2x}+1$

18. wio

So it isn't this?$\Large \int_0^1 \frac{e^2}{e^{2x}+1} dx$

19. wio

$\Large \int_0^1 \frac{e^x}{e^{2x}+1} dx$

20. jotopia34

yes

21. satellite73

think of $\int\frac{x}{1+x^2}dx$

22. Goten77

|dw:1359689203929:dw| eh i think im doing something wrong....

23. Goten77

and then satellite makes it look ez <.<

24. jotopia34

wait, I thought you can't have a u and an x in the equation

25. Goten77

thats right jotopia... thats y i knew i did it wrong XD

26. satellite73

$u=e^x,du=e^xdx$ you get $\int\frac{du}{1+u^2}$

27. jotopia34

I don't see what sattelite said because how can e^2x be like x^2

28. satellite73

oh it is a definite integral

29. jotopia34

This is scandalous!!

30. satellite73

$$u(0)=1,u(1)=e$$ you get $\int_1^e\frac{du}{1+u^2}$

31. wio

Since $$u=e^{2x}=(e^x)^2 \implies \sqrt{u}=e^x$$, @Goten77 's method isn't hopeless.

32. satellite73

anti derivative of $$\frac{1}{1+u^2}$$ is $$\tan^{-1}(u)$$

33. Goten77

this is 1 of those like situations u gotta memorize.... its um arc tan i believe? im not sure.......been a while since i done this stuff

34. jotopia34

oh, e^2x is the same as saying e^x squared????/

35. satellite73

yes it certainly is

36. jotopia34

like how cos^2x is (cosx)^2

37. satellite73

$$(e^x)^2=e^{2x}$$

38. satellite73

laws 'o exponents

39. jotopia34

it amazes me that no one ever pointed that out before

40. jotopia34

lol. Laws 'o fuh shizzle

41. satellite73

and also you are right, $$\cos^2(x)=(\cos(x))^2$$

42. satellite73

it is just more common to write $$f^2(x)$$ than $$(f(x))^2$$

43. jotopia34

yes, but its confusing because if you said (e^x)^2, would it be like saying $e ^{2}x ^{2}$

44. wio

You're confusing multiplication with exponents

45. Goten77

jotopia look at this my pic below

46. jotopia34

hmmm, but isnt (2x)^2 the same as 2^2x^2. And e is a number

47. wio

$$(ex)^2 = e^2x^2$$ but $$(e^x)^2 =e^{2x}$$

48. Goten77

|dw:1359689988230:dw|

49. jotopia34

oh yeah!! I get it now

50. jotopia34

that is key. You are all fantabulous

51. wio

Anyway, if you sub $$u = e^{2x}+1$$ then $$e^x = \sqrt{u^2-1}$$ which is also complicated

52. jotopia34

damn, nope I am still confused because if i asked you to take the derivative of $(e ^{x})^2$, you would do $2(e^x)(e^x) = 2(e^x)^2=2e ^{2x}$. Yep i get it

53. jotopia34

damn!! I still don't get it $u=e^{2x}, du=2e^{2x}dx$ =$\int\limits_{0}^{1} \frac{ e^x}{ 1+u }* \frac{ du }{ 2u }$, lost!!

54. wio

Anyway, satelitte73's answer is correct, but if you didn't memorize that $$d/dx \tan^-{1}(x)=1/(x^2+1)$$ there is a way to do it with trig subs

55. jotopia34

so, no u sub?

56. wio

Well you two two substitutions:$1)\quad u= e^x \\ 2)\quad u=\tan(\theta)$

57. jotopia34

is the answer tan^-1e^x$\tan^-1e^x$

58. wio

That's the anti-derivative, you have to put in the limits.

59. jotopia34

yes you are right but is that the antiderivative part?

60. wio

Yeah, though I'd like to see your work.

61. Hoa

Hi friend, you have a formula for the problem, why don't you apply? let u =e^x ---> du = e^x dx. you have the new numerator is du, in denominator, you have u^2 +1 . and then change the interval : when x =0 , u =1, when x =1, u =e, so your new one is integral from 1 to e of du/u^2 + 1 = arctan (u) from 0 to 1 = arctan1 - arctan 0 . arctan 1 = pi/4 and arctan 0 = 0 . so the final result is pi/4. is it familiar with you? Hope this help

62. Hoa

oh I'm sorry, arctan (e) - arctan 1. not as above. terribly sorry