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SinginDaCalc2Blues

  • one year ago

Hi! Integrate 2x/(x^8+1) dx

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  1. primeralph
    • one year ago
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    I'll be honest, I can solve that......the solution is not neat......so don't expect anyone here to be able to finish it....it's long...........try wolfram....... but you still can try here though

  2. SinginDaCalc2Blues
    • one year ago
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    can you tell me how to at least start it?

  3. smokeydabear
    • one year ago
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    I feel like this will end up being an ln function, but I think there's a lot more to it than just that I'm sure. Let Ralph help you.

  4. SinginDaCalc2Blues
    • one year ago
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    does anyone know of a completed example of a problem like this I can go online and review or anything?

  5. primeralph
    • one year ago
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    |dw:1368859626389:dw|

  6. primeralph
    • one year ago
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    That's how you start.

  7. SinginDaCalc2Blues
    • one year ago
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    can you tell me what the J is for? I have never ever seen that used before

  8. primeralph
    • one year ago
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    |dw:1368859717310:dw|

  9. primeralph
    • one year ago
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    J is just a multiplier. In this case, J = 1 , so you can remove it

  10. felavin
    • one year ago
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    (x^8+1) means: X^9

  11. smokeydabear
    • one year ago
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    I highly doubt that it would be that simple felavin.

  12. felavin
    • one year ago
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    WHY? SMOKEYDABEAR.

  13. SinginDaCalc2Blues
    • one year ago
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    could the next step be to let u=tan(theta) then du=sec^2(theta) d(theta)

  14. primeralph
    • one year ago
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    @SinginDaCalc2Blues same thing.........you're looping

  15. smokeydabear
    • one year ago
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    Yeah all that would do is put you right back to where you started

  16. primeralph
    • one year ago
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    ^^^ that girl is just too smart..........

  17. smokeydabear
    • one year ago
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    Too stupid*

  18. smokeydabear
    • one year ago
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    I'm just here to learn how to do it myself

  19. SinginDaCalc2Blues
    • one year ago
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    are you people calling me stupid? seriously?

  20. smokeydabear
    • one year ago
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    No, I was calling myself stupid xD

  21. primeralph
    • one year ago
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    no, I didn't mean that.

  22. primeralph
    • one year ago
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    okay, let's figure out an easier way to solve this problem.

  23. primeralph
    • one year ago
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    @SinginDaCalc2Blues can you handle recursive integrals?

  24. SinginDaCalc2Blues
    • one year ago
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    okay...well I appreciate the help from everyone but it's hard to understand the context of non-verbal communication sometimes...I was hoping you guys weren't calling me stupid...anyway...

  25. primeralph
    • one year ago
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    @SinginDaCalc2Blues we really would never do that.

  26. SinginDaCalc2Blues
    • one year ago
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    i have never heard the term recursive integral...while I probably know what your talking about, I don't know it in those terms...can you describe?

  27. primeralph
    • one year ago
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    wait............a minute...why not just polar?

  28. smokeydabear
    • one year ago
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    Polar coordinates? Actually that makes some sense...

  29. SinginDaCalc2Blues
    • one year ago
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    are you saying turn the integral into a polar expression or something?

  30. primeralph
    • one year ago
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    @smokeydabear tricky thing is, it returns the integrand to the form we had before....because using cos, cos = 1/sec.....

  31. SinginDaCalc2Blues
    • one year ago
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    my professor is devious!

  32. smokeydabear
    • one year ago
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    Your professor is pretty evil for giving you something like this

  33. Peter14
    • one year ago
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    would u=x^2 du=2x dx indefinite integral of 1/(u^4+1) help? (sorry, my latex isn't working)

  34. primeralph
    • one year ago
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    @SinginDaCalc2Blues I reduced it to......can you solve this?|dw:1368860855998:dw|

  35. felavin
    • one year ago
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    the int, called complex in

  36. SinginDaCalc2Blues
    • one year ago
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    uhm if k is a constant, what is J again?

  37. primeralph
    • one year ago
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    J is the variable of integration, just like U

  38. primeralph
    • one year ago
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    U in U substitution

  39. smokeydabear
    • one year ago
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    I just looked it up on wolfram... what is this monstrosity...

  40. primeralph
    • one year ago
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    @smokeydabear I told you...........

  41. smokeydabear
    • one year ago
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    How do you know where to even start with something like this? That's always been my problem

  42. SinginDaCalc2Blues
    • one year ago
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    a good place to start is with Vodka :o)

  43. primeralph
    • one year ago
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    @SinginDaCalc2Blues I asked you if you knew recursives because as you can see here, the trigs are repeating..which means recursion

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  44. Peter14
    • one year ago
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    no, not vodka, start with despair.

  45. SinginDaCalc2Blues
    • one year ago
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    in 30 minutes i will no longer be able to do integration but I will be able to slurrrr Russian! :o)~

  46. Peter14
    • one year ago
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    vodka inhibits your ability to do math.

  47. Peter14
    • one year ago
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    despair only temporarily inhibits your ability to do everything

  48. primeralph
    • one year ago
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    @smokeydabear recursion works to start. for example|dw:1368861204284:dw|

  49. smokeydabear
    • one year ago
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    Okay, so how would we use that to help us? I'm still confused :s

  50. some_someone
    • one year ago
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    crazy problem o:

  51. some_someone
    • one year ago
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    well long only :P

  52. primeralph
    • one year ago
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    @smokeydabear it saves time because once you know that it will repeat, you can generalize..........

  53. SinginDaCalc2Blues
    • one year ago
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    hey guys...i have an idea...gimme a sec...

  54. primeralph
    • one year ago
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    .........meaning............|dw:1368861457508:dw|

  55. stamp
    • one year ago
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    u = x^2, du = 2x dx\[\int \frac{du}{u^4+1}\]

  56. primeralph
    • one year ago
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    so, because they occur in trig functions, we can use it for the tan theta

  57. primeralph
    • one year ago
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    @stamp did that

  58. smokeydabear
    • one year ago
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    Ohhhhhh I see what you did with the sin's and the pi's @primeralph very clever! =D

  59. Peter14
    • one year ago
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    now we just need to learn to use it effectively...

  60. primeralph
    • one year ago
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    |dw:1368861979382:dw|

  61. smokeydabear
    • one year ago
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    I gotcha Ralph, I noticed it too but I didn't want to be mean =P

  62. SinginDaCalc2Blues
    • one year ago
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  63. SinginDaCalc2Blues
    • one year ago
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    Does this little bit of trickyness open any doors?

  64. SinginDaCalc2Blues
    • one year ago
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    yeah what stamp wrote I believe is wrong...it wouldn't be x^4...it would be (x^4)^2

  65. smokeydabear
    • one year ago
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    Well what stamp wrote isn't technically wrong, but it isn't the method to solving this.

  66. SinginDaCalc2Blues
    • one year ago
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    I basically factored out of the denominator an x^2...then did u-sub

  67. SinginDaCalc2Blues
    • one year ago
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  68. felavin
    • one year ago
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    hi, the integrate is simple: Who can solve this integrate:\[\int\limits_{}^{} \frac{\tan Z }{ Z- \frac{ P }{ 4}}\] GAME IS OVER ..........

  69. felavin
    • one year ago
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    hello

  70. SinginDaCalc2Blues
    • one year ago
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    hi...i'm only in calc 2...i don't know what Cauchy-Reimann equations are... I did however get my integral down to 1/(u^3+1)

  71. drawar
    • one year ago
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    In case you love twisting with partial fractions: \[\frac{ 2x }{ x^8 +1}=\frac{ 2x }{( x^4-\sqrt{2}x^2+1)( x^4+\sqrt{2}x^2+1)}\]

  72. SinginDaCalc2Blues
    • one year ago
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    hi drawer...i think your way is looking promising...i think i was headed toward that...can you look at my .bmp and tell me how to start partial fraction with what i ended with?

  73. SinginDaCalc2Blues
    • one year ago
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  74. drawar
    • one year ago
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    Then \[u^3+1=(u+1)(u^2-u+1)\], use partial fractions.

  75. SinginDaCalc2Blues
    • one year ago
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    what do i do with the numerator?

  76. drawar
    • one year ago
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    \[\frac{ 1 }{ u^3+1 }=\frac{ A }{ u+1 }+\frac{ Bu+C }{ u^2-u+1 }\] =>\[A(u^2-u+1)+(Bu+C)(u+1)=1\] Then equate the coefficients and you'll get a system of linear equations in terms of A,B, and C. Solve it!

  77. SinginDaCalc2Blues
    • one year ago
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    so then...after I do that, will i have technically several different integrals which will be easier to integrate?

  78. drawar
    • one year ago
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    Yes, you'll see that the degree of the polynomial of the numerator is always 1 less than that of the denominator, which must have sth to do with the natural logarithm (ln) function.

  79. shubhamsrg
    • one year ago
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    i havent read much what is written above, its too long, has the question been solved ?

  80. SinginDaCalc2Blues
    • one year ago
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    ok drawar...the answer definitely has some nasty ln's...i will keep plugging away at it...one more question...

  81. SinginDaCalc2Blues
    • one year ago
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    once i figure out what A, B, AND C equate to, don't I just plug those into the numerators and then treat one as a separate integral to solve?

  82. shubhamsrg
    • one year ago
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    has anyone tried x^2 =t ?

  83. SinginDaCalc2Blues
    • one year ago
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    yeah shub...wayyy up at the top...we are past that now...but thanks

  84. drawar
    • one year ago
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    Yes @SinginDaCalc2Blues

  85. shubhamsrg
    • one year ago
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    well, why you didnt go forward with that method, after that substitution, we'll have dt/(t^4 +1) note that t^4 + 1 = (t^2 + i) (t^2 -i) also note that (t^2 +i) - ( t^ -i) = 2i , a constant

  86. SinginDaCalc2Blues
    • one year ago
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    ok...i will keep plugging away and see if I can get it...this problem has been so crazy! :o)

  87. SinginDaCalc2Blues
    • one year ago
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    are you using imaginary numbers shub?

  88. drawar
    • one year ago
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    is it a definite/indefinite integral in your original question?

  89. shubhamsrg
    • one year ago
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    yes, the very same. i can be treated as just another constant

  90. SinginDaCalc2Blues
    • one year ago
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    indefinite

  91. SinginDaCalc2Blues
    • one year ago
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    i have no idea how to use imaginary numbers...i'm only in calc 2..we haven't used those

  92. shubhamsrg
    • one year ago
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    I see, well use of iota will lead you to the solution withing 2 more steps, and in the end you can always replace the complex part with hyperbolic functions, or natural logarithms . But anyways, since you havent learnt about that, go with @drawar 's method (:

  93. SinginDaCalc2Blues
    • one year ago
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    I wish I had learned about them...i just hope i can solve this

  94. RolyPoly
    • one year ago
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    \[\int\frac{2x}{x^8+1}dx\]\[=\int \frac{1}{x^8+1}d(x^2)\]\[=\int \frac{1}{y^4+1}dy\]\[=\frac{1}{2}\int \frac{y^2+1-y^2 +1}{y^4+1} dy\]\[=\frac{1}{2}(\int \frac{y^2+1}{y^4+1}dy-\int\frac{y^2 -1}{y^4+1} dy)\]\[=\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\int\frac{1 -\frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\] Consider \[\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\] Let u = \(y-\frac{1}{y}\) \(du =(1 + \frac{1}{y^2}) dy\) and \(y^2 + 1/y^2 = (y-\frac{1}{y})^2+2=u^2+2\) So, \[\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\]\[=\int \frac{1}{u^2+2}du\] Similarly, for\(\int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\) Let \(v = y+ \frac{1}{y}\) \[dv = 1 - \frac{1}{y^2}dy\]\[y^2 + 1/y^2 = (y+\frac{1}{y})^2-2=v^2-2\] So, \[\int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\]\[=\int\frac{1}{v^2-2}dv\] As a result, \[\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\int\frac{1 -\frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\]\[=\frac{1}{2}(\int\frac{1}{u^2+2}du-\int\frac{1}{v^2-2}dv)\]And You can integrate it then.

  95. primeralph
    • one year ago
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    @roly Thank you for taking the time to illustrate it.

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