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anonymous

  • one year ago

Okay, so if you don't like math then don't try to answer this. I've spent the last hour doing it. ∫3/(x^3−1)dx

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  1. anonymous
    • one year ago
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    what's the question?

  2. anonymous
    • one year ago
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    \[\int\limits_{}^{}3/(x^3-1)dx\]

  3. anonymous
    • one year ago
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    is the '/' on purpose?

  4. anonymous
    • one year ago
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    Yeah, thats division.

  5. anonymous
    • one year ago
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    oh ok

  6. anonymous
    • one year ago
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    I get lost when I'm at log|x-1|-\[\int\limits_{}^{}\frac{ x+2 }{ x^2+x+1}\]

  7. anonymous
    • one year ago
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    ln|x-1|*

  8. jtvatsim
    • one year ago
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    So you used partial fractions to get to this point correct?

  9. anonymous
    • one year ago
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    Yeah.

  10. jtvatsim
    • one year ago
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    Alright, so let me double check that on my end.

  11. anonymous
    • one year ago
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    I got

  12. jtvatsim
    • one year ago
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    Cool. I checked over here and it looks correct so far.

  13. jtvatsim
    • one year ago
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    Still thinking here... obviously, this is a tricky one. :)

  14. amistre64
    • one year ago
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    what part is losing you?

  15. amistre64
    • one year ago
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    \[\frac{ x+2 }{ x^2+x+1}\] \[\frac{ (x+1)+1 }{ x^2+x+1}\] what is the derivative of the bottom?

  16. anonymous
    • one year ago
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    Let u=x^2+x+1 and du=2x+1dx, \[\frac{ x+2 }{ x^2+x+1 }=\frac{ 1 }{ 2 }*\frac{ 2x+1 }{ x^2+x+1 }+\frac{ 3 }{ 2 }*\frac{ 1 }{ x^2+x+1 }\]

  17. anonymous
    • one year ago
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    I've gotten about up to that point, so you can say \[\int\limits_{}^{}\frac{ 1 }{ 2 }*\frac{ du }{ u }=\frac{ 1 }{ 2 }\ln|u|+C=\frac{ 1 }{ 2 }\ln|x^2+x+1|+C\]

  18. amistre64
    • one year ago
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    fine, we can do that too :) \[\frac12~\frac{ 2(x+1) }{ x^2+x+1}\] \[\frac12~\frac{ 2x+2) }{ x^2+x+1}\] \[\frac12~\frac{ (2x+1)+1 }{ x^2+x+1}\]

  19. anonymous
    • one year ago
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    But then \[\frac{ 3 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ x^2+x+1 }dx\]

  20. amistre64
    • one year ago
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    completing the square on the bottom might be useful ... been awhile tho

  21. amistre64
    • one year ago
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    the other option is to just decompose it over the complex plane

  22. anonymous
    • one year ago
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    Oh okay, so its \[\frac{ 1 }{ (x+.5)^2+.75 }\] ? and now I'm pretty lost. It looks like maybe you can do some trig substitution but I can't really tell.

  23. amistre64
    • one year ago
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    what are your inverse trig derivatives? reviewing them might help out ... tan^-1 rings a bell

  24. anonymous
    • one year ago
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    Dont wanna type them all out so : http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns_files/eq0044M.gif

  25. amistre64
    • one year ago
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    y = tan^-1 (x) tan(y) = x y' sec^2(y) = 1 y' = 1/sec^2(y) y' = 1/(tan^2(y)+1) y' = 1/(tan^2(tan^-1(x))+1) y' = 1/(x^2+1)

  26. amistre64
    • one year ago
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    typing them out helps to keep them in memory .... pauls site wont always be available :)

  27. anonymous
    • one year ago
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    True :P

  28. anonymous
    • one year ago
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    \[\int\limits_{}^{}\frac{ 1 }{ (x+.5)^2+.75 }dx=\int\limits_{}^{}\frac{ \sqrt3/2*\sec^2(\theta) }{ 3/4*\tan^2\theta+1 }\]

  29. anonymous
    • one year ago
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    I think.... ._.

  30. amistre64
    • one year ago
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    im pretty sure im making a mess lol, but heres my thought process if we go this archaic route 1 -------------- (x+1/2)^2 + 3/4 4/3 ---------------- 4/3(x+1/2)^2 + 1 4/3 ---------------- (16/9 (x+1/2))^2 + 1 assuming we can work out some tan inverse y = K tan^-1(A(x+B)) y/K = tan^-1(A(x+B)) tan(y/K) = A(x+B) y' sec^2(y/K)/K = A y' sec^2(y/K) = KA y' (tan^2(y/K)+1) = KA y' ((A(x+B))^2+1) = KA y' = KA ------------ (A(x+B))^2+1 ------------------- A = 16/9, B=1/2 16K/9 = 4/3 16K = 12 K = 3/4

  31. anonymous
    • one year ago
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    Omg, \[\tan^2\theta+1=\sec^2\theta, so \it simplifies \to \int\limits_{}^{}\frac{ 2 }{ \sqrt3 }d \theta\]

  32. amistre64
    • one year ago
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    you are on the right track yes http://www.wolframalpha.com/input/?i=integrate+1%2F%28x^2%2Bx%2B1%29+dx

  33. amistre64
    • one year ago
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    mine went someplace off the wall lol

  34. amistre64
    • one year ago
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    4/3 sqrts to get inside the ^2 ...

  35. amistre64
    • one year ago
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    2/sqrt(3) not 16/9

  36. amistre64
    • one year ago
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    A = 2/sqrt(3) 2K/sqrt(3) = 4/3 2sqrt(3) K = 4 K = 2/sqrt(3) thats better on my end

  37. anonymous
    • one year ago
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    \[= \frac{ 2 }{ \sqrt3 } \theta+C, \theta=\arctan(\frac{ 2 }{ \sqrt3 }(x+.5)) \implies \frac{ 2 }{ \sqrt3 }\arctan(\frac{ 2 }{ \sqrt3}(x+.5))+C=\int\limits_{}^{}\frac{ 1 }{ x^2+x+1 }dx\]

  38. amistre64
    • one year ago
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    now me and the wolf agree: y = K tan^-1 (A(x+B)) \[y=\frac{2}{\sqrt3}~\tan^{-1}[\frac{2}{\sqrt3}(x+\frac12)]\]

  39. anonymous
    • one year ago
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    This answer is going to be really ugly in the end

  40. anonymous
    • one year ago
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    Yeah, I agree, then we just put it all together and its \[\log|x-1|-0.5\log|x^2+x+1|-\sqrt3\arctan(\frac{ 2 }{ \sqrt3 }(x+.5)+C\] I dont think that simplifies any further

  41. amistre64
    • one year ago
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    the uglier the better :)

  42. amistre64
    • one year ago
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    you got it

  43. anonymous
    • one year ago
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    Jeez, that was tough.

  44. amistre64
    • one year ago
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    nah .. it was "interesting" lol

  45. anonymous
    • one year ago
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    Lol

  46. amistre64
    • one year ago
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    good luck, its supper time

  47. anonymous
    • one year ago
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    Thank you btw.

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