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wasiqss

  • 3 years ago

Integral of (e^x)/x

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  1. waterineyes
    • 3 years ago
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    Applying the Divide rule of Derivative here..

  2. wasiqss
    • 3 years ago
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    lol water do it by that way i will give you thousand medals :P

  3. wasiqss
    • 3 years ago
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    and its integration man~

  4. waterineyes
    • 3 years ago
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    Sorry I did not read the question fully..

  5. Omniscience
    • 3 years ago
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    \[\large \int\limits\frac{e^{x}}{x} dx\]?? looks like cant be solved by elementary methods..

  6. wasiqss
    • 3 years ago
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    ahaha omni in your dreams do it :P

  7. waterineyes
    • 3 years ago
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    We can use here Integration by Parts...

  8. Omniscience
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integrate+%28e^x%29%2Fx like srsly? how can you solve it?

  9. wasiqss
    • 3 years ago
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    hahahah omni i heard you saying i own calculus ? :P

  10. Omniscience
    • 3 years ago
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    term by term integration of the infinite series?

  11. Omniscience
    • 3 years ago
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    like srsly.. http://en.wikipedia.org/wiki/Exponential_integral

  12. meera_yadav
    • 3 years ago
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    hi water , have you heard of exp x = \[x + x ^{2}/2! + x ^{3}/3! + x ^{4}/4!.........\]

  13. Outkast3r09
    • 3 years ago
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    Wasi, what calculus are you in

  14. meera_yadav
    • 3 years ago
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    if you know this then u can substitute exp x by the above and further see if its solvable !

  15. wasiqss
    • 3 years ago
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    keep thinking :P

  16. wasiqss
    • 3 years ago
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    hint it can be done by parts

  17. Outkast3r09
    • 3 years ago
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    i know it can be done by parts however are you in calculus 2

  18. wasiqss
    • 3 years ago
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    well outkas do it by parts i wanna see

  19. waterineyes
    • 3 years ago
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    For convenience, 1/x is the first function.. e^x as second function.. \[\int\limits_{}^{}(\frac{1}{x}).e^x.dx = \frac{1}{x}.e^x - \int\limits_{}^{}(\frac{-1}{x^2}.e^x).dx\] \[= \frac{e^x}{x} +\int\limits_{}^{} \frac{e^x}{x^2}.dx\] But now x will move on increasing like x^3, x^4 etc etc..

  20. wasiqss
    • 3 years ago
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    yeah now you have to do integration of (e^x)/x^2 right?

  21. Outkast3r09
    • 3 years ago
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    yes however switch the u and dv to go back and then add them and divide by 2

  22. waterineyes
    • 3 years ago
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    Right but that will lead to x^3 then this process continues..

  23. Outkast3r09
    • 3 years ago
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    in otherword let dv = -1/x^2

  24. wasiqss
    • 3 years ago
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    haha outkas right :D :D

  25. Outkast3r09
    • 3 years ago
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    then add the integrals and divide by 2

  26. wasiqss
    • 3 years ago
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    it is a two step integral

  27. wasiqss
    • 3 years ago
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    @Omniscience poor you :P

  28. wasiqss
    • 3 years ago
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    outkas which calculus you doing

  29. Outkast3r09
    • 3 years ago
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    I'm don with calculus, I'm in Differential Equations right now

  30. wasiqss
    • 3 years ago
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    im good at D.E's :D

  31. Outkast3r09
    • 3 years ago
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    however i don't remember hardly anything from calc 3.. and i wouldn't remember the series i it weren't for review

  32. wasiqss
    • 3 years ago
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    calc 3 sucks!

  33. Outkast3r09
    • 3 years ago
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    indeed it does

  34. wasiqss
    • 3 years ago
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    D.E are real good, in which uni you are and which year?

  35. Outkast3r09
    • 3 years ago
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    nothing like drawing 3d objects on a 2d piece of paper

  36. Outkast3r09
    • 3 years ago
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    i'm a at a community and it's around my second/3rd year

  37. wasiqss
    • 3 years ago
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    i just cleared the first year only :/

  38. wasiqss
    • 3 years ago
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    now i will go to second

  39. meera_yadav
    • 3 years ago
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    good try mukushla :)

  40. mukushla
    • 3 years ago
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    tnx

  41. mukushla
    • 3 years ago
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    there was a little mistake in my answer \[\int\limits \frac{e^x}{x}dx=\int\limits \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^n}{n!}dx=\sum_{n=0}^{\infty} \int\limits \frac{x^{n-1}}{n!} dx=\ln x+\sum_{n=1}^{\infty} \frac{x^{n}}{n.n!}\]

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