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jotopia34

  • 2 years ago

Can anyone help me with the Integral of e^x/(1+e^x) dx

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  1. Goten77
    • 2 years ago
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    |dw:1359688972018:dw|

  2. Goten77
    • 2 years ago
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    hmm

  3. jotopia34
    • 2 years ago
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    \[\int\limits_{0}^{1}e^x/e ^{2x}+1\]

  4. jotopia34
    • 2 years ago
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    sorry I made a mistake its e^2x

  5. Goten77
    • 2 years ago
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    oh then

  6. wio
    • 2 years ago
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    Try u sub

  7. wio
    • 2 years ago
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    \(u=e^x+1\)

  8. jotopia34
    • 2 years ago
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    but it I do u=e^2x, du= 2e^2x and I still can't get rid of the e^x

  9. wio
    • 2 years ago
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    Okay, what exactly is the equation?

  10. Goten77
    • 2 years ago
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    |dw:1359689045480:dw|

  11. jotopia34
    • 2 years ago
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    nope. I still dont have it.....

  12. wio
    • 2 years ago
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    What is the equation?

  13. wio
    • 2 years ago
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    I do not know what the correct equation is.

  14. jotopia34
    • 2 years ago
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    the euation is above

  15. wio
    • 2 years ago
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    There are MANY equations above.

  16. wio
    • 2 years ago
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    Which one is correct?

  17. jotopia34
    • 2 years ago
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    \[\int\limits_{0}^{1}e^x/e ^{2x}+1\]

  18. wio
    • 2 years ago
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    So it isn't this?\[\Large \int_0^1 \frac{e^2}{e^{2x}+1} dx \]

  19. wio
    • 2 years ago
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    \[ \Large \int_0^1 \frac{e^x}{e^{2x}+1} dx \]

  20. jotopia34
    • 2 years ago
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    yes

  21. satellite73
    • 2 years ago
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    think of \[\int\frac{x}{1+x^2}dx\]

  22. Goten77
    • 2 years ago
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    |dw:1359689203929:dw| eh i think im doing something wrong....

  23. Goten77
    • 2 years ago
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    and then satellite makes it look ez <.<

  24. jotopia34
    • 2 years ago
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    wait, I thought you can't have a u and an x in the equation

  25. Goten77
    • 2 years ago
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    thats right jotopia... thats y i knew i did it wrong XD

  26. satellite73
    • 2 years ago
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    \[u=e^x,du=e^xdx\] you get \[\int\frac{du}{1+u^2}\]

  27. jotopia34
    • 2 years ago
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    I don't see what sattelite said because how can e^2x be like x^2

  28. satellite73
    • 2 years ago
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    oh it is a definite integral

  29. jotopia34
    • 2 years ago
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    This is scandalous!!

  30. satellite73
    • 2 years ago
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    \(u(0)=1,u(1)=e\) you get \[\int_1^e\frac{du}{1+u^2}\]

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

  32. satellite73
    • 2 years ago
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    anti derivative of \(\frac{1}{1+u^2}\) is \(\tan^{-1}(u)\)

  33. Goten77
    • 2 years ago
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    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
    • 2 years ago
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    oh, e^2x is the same as saying e^x squared????/

  35. satellite73
    • 2 years ago
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    yes it certainly is

  36. jotopia34
    • 2 years ago
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    like how cos^2x is (cosx)^2

  37. satellite73
    • 2 years ago
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    \((e^x)^2=e^{2x}\)

  38. satellite73
    • 2 years ago
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    laws 'o exponents

  39. jotopia34
    • 2 years ago
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    it amazes me that no one ever pointed that out before

  40. jotopia34
    • 2 years ago
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    lol. Laws 'o fuh shizzle

  41. satellite73
    • 2 years ago
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    and also you are right, \(\cos^2(x)=(\cos(x))^2\)

  42. satellite73
    • 2 years ago
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    it is just more common to write \(f^2(x)\) than \((f(x))^2\)

  43. jotopia34
    • 2 years ago
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    yes, but its confusing because if you said (e^x)^2, would it be like saying \[e ^{2}x ^{2}\]

  44. wio
    • 2 years ago
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    You're confusing multiplication with exponents

  45. Goten77
    • 2 years ago
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    jotopia look at this my pic below

  46. jotopia34
    • 2 years ago
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    hmmm, but isnt (2x)^2 the same as 2^2x^2. And e is a number

  47. wio
    • 2 years ago
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    \((ex)^2 = e^2x^2\) but \((e^x)^2 =e^{2x}\)

  48. Goten77
    • 2 years ago
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    |dw:1359689988230:dw|

  49. jotopia34
    • 2 years ago
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    oh yeah!! I get it now

  50. jotopia34
    • 2 years ago
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    that is key. You are all fantabulous

  51. wio
    • 2 years ago
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    Anyway, if you sub \(u = e^{2x}+1\) then \(e^x = \sqrt{u^2-1}\) which is also complicated

  52. jotopia34
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    so, no u sub?

  56. wio
    • 2 years ago
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    Well you two two substitutions:\[ 1)\quad u= e^x \\ 2)\quad u=\tan(\theta) \]

  57. jotopia34
    • 2 years ago
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    is the answer tan^-1e^x\[\tan^-1e^x\]

  58. wio
    • 2 years ago
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    That's the anti-derivative, you have to put in the limits.

  59. jotopia34
    • 2 years ago
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    yes you are right but is that the antiderivative part?

  60. wio
    • 2 years ago
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    Yeah, though I'd like to see your work.

  61. Hoa
    • 2 years ago
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    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
    • 2 years ago
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    oh I'm sorry, arctan (e) - arctan 1. not as above. terribly sorry

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