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anonymous

  • 5 years ago

integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^2-15)) should i do partial fraction 1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^2-15)) = A / x + B/x^2 + Cx+d / ( x^2 - 15)

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  1. amistre64
    • 5 years ago
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    oy... if you hit the enter key it spreads this out to where it is more readable.... just a thought

  2. anonymous
    • 5 years ago
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    the x^2 - 15 is reducible , but im not sure if youre supposed to use that theorem

  3. anonymous
    • 5 years ago
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    enter key?

  4. amistre64
    • 5 years ago
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    the key on the keyboard that makes this happen ;)

  5. anonymous
    • 5 years ago
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    you want me to use the enter key more often?

  6. amistre64
    • 5 years ago
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    yes, difference of squares that part and you should have 3 sets to figure out

  7. anonymous
    • 5 years ago
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    but you get radicals

  8. amistre64
    • 5 years ago
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    radicals are fine.... right?

  9. anonymous
    • 5 years ago
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    are they ? how come the other way works just as well?

  10. amistre64
    • 5 years ago
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    which other way can you decompose the fraction?

  11. anonymous
    • 5 years ago
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    1/((x^2*(x^2-15)) = A / x + B/x^2 + (Cx+d) / ( x^2 - 15)

  12. anonymous
    • 5 years ago
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    technically the quadratic is reducible

  13. amistre64
    • 5 years ago
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    it could be a fluke if it does work out....

  14. amistre64
    • 5 years ago
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    or it could be a possibility for solutions that i am unaware of as yet :)

  15. anonymous
    • 5 years ago
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    no it isnt a fluke, in general one need not reduce a quadratic

  16. amistre64
    • 5 years ago
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    in general you should, to account for multiples i think

  17. amistre64
    • 5 years ago
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    if it aint got multiples, then you should be fine i spose

  18. amistre64
    • 5 years ago
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    right?

  19. anonymous
    • 5 years ago
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    yes it usually makes it easier to reduce the quadratic, except in the case where there are radical linear factors

  20. anonymous
    • 5 years ago
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    it can have multiples, theres no problem as far as i can see

  21. amistre64
    • 5 years ago
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    without doing a ton of them to generalize it, id have to throw it in the hmmmmm box lol

  22. anonymous
    • 5 years ago
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    say integral ( 1 / (x(x+1)^2) = A/x + (Bx+C) / (x+1)^2

  23. anonymous
    • 5 years ago
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    its a little weird, might be easier if i wrote 1/( x ( x^2+2x+1)) = A/x + (Bx+c) / (x^2 + 2x + 1)

  24. anonymous
    • 5 years ago
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    sometimes it is difficult to factor a quadratic, right?

  25. anonymous
    • 5 years ago
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    ok , i should get into more accurate terms. sometimes it is reducible over reals, not rationals

  26. amistre64
    • 5 years ago
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    factoring a quadratic is fairly easy; might get complicated solutions, but farily easy to do

  27. anonymous
    • 5 years ago
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    yes but doing partial fractions with irrational roots is tough

  28. amistre64
    • 5 years ago
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    and i have heard that you can use the complex solutions as well with no ill effects on the decomp

  29. anonymous
    • 5 years ago
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    ok , so youre so confident, try to do parfractions on 1 / ( x^2(x^2-15) , give it a go

  30. anonymous
    • 5 years ago
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    i was stuck

  31. anonymous
    • 5 years ago
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    1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15)

  32. amistre64
    • 5 years ago
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    confident? lol ; just stubborn

  33. anonymous
    • 5 years ago
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    it can even get stranger, this is an easy irrational solution problem, you can have say A/ ( x - sqrt 3/2 ) for instance

  34. anonymous
    • 5 years ago
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    this is why calculus in textbooks is generally contrived , or idealized problems. in the real world you have these irrational roots, etc

  35. anonymous
    • 5 years ago
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    theorem time, one sec

  36. amistre64
    • 5 years ago
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    1 a b c d ----------- = --- + ----- + --------- + --------- x^2(x^2-15) x x^2 x-sqrt(15) x+sqrt(15) is the 'proper' set up right?

  37. anonymous
    • 5 years ago
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    yes looks good

  38. anonymous
    • 5 years ago
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    Personally, I'd leave the last fraction's denominator without doing difference of two squares.

  39. anonymous
    • 5 years ago
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    1 / [(x-a)^n (x^2+bx+c)^n] = A1/(x-a) + A2/ ( x-a)^2 + ... An/(x-a)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

  40. anonymous
    • 5 years ago
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    so the question is, do the quadratics have to be irreducible, i dont think so

  41. amistre64
    • 5 years ago
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    1 a b c d ----------- = --- + ----- + --------- + --------- x^2(x^2-15) x x^2 x-sqrt(15) x+sqrt(15) a(x^2)(x-sqrt(15))(x+sqrt(15))) b(x)(x-sqrt(15))(x+sqrt(15))) c(x)( x^2)(x+sqrt(15)) d(x)(x^2)(x-sqrt(15))

  42. anonymous
    • 5 years ago
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    MOST calculus books and websites say the quadratic is irreducible though

  43. anonymous
    • 5 years ago
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    irreducible over the reals i imagine. wikipedia is better, they say irreducible over the field

  44. anonymous
    • 5 years ago
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    whatever field you have in mine

  45. anonymous
    • 5 years ago
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    For this particular problem: \[\frac{1}{y(y-15)} \equiv \frac{1}{15(y-15)} - \frac{1}{15y}\] \[\text{Let } y = x^2\] ...

  46. anonymous
    • 5 years ago
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    newton, be careful though

  47. anonymous
    • 5 years ago
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    if you are substituting a variable, you will need to do 2x dx = dy

  48. anonymous
    • 5 years ago
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    so you might want to just change it back. thats an identity

  49. anonymous
    • 5 years ago
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    i should write a book on calculus pitfalls, since i keep finding them :)

  50. anonymous
    • 5 years ago
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    newtons, so just change y back to x^2

  51. anonymous
    • 5 years ago
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    Sorry, in case it wasn't clear, it leads to: \[\frac{1}{x^2(x^2-15)} \equiv \frac{1}{15(x^2-15)} + \frac{1}{15x^2} \] I would integrate this.

  52. anonymous
    • 5 years ago
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    otherwise you have 2x dx = dy , dx = dy / ( 2x ) = dy / ( 2 sqrt y )

  53. anonymous
    • 5 years ago
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    correct, i thought you were doing a change of variables

  54. amistre64
    • 5 years ago
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    1 = a(x^2)(x-sqrt(15))(x+sqrt(15))) + b(x)(x-sqrt(15))(x+sqrt(15))) + c(x)( x^2)(x+sqrt(15)) +d(x)(x^2)(x-sqrt(15)) when x = sqrt(15) we get 1 = c(sqrt(15))(15)(2sqrt(15)) 1 = c(15)(2)(15) 1 = c(225)(2) = c450 c = 1/450

  55. anonymous
    • 5 years ago
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    amistre, you get medal for persistence :)

  56. anonymous
    • 5 years ago
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    but its much easier if we dont split the quadratic

  57. anonymous
    • 5 years ago
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    I was only doing that to show that you don't need a 1/x fraction and 1/x^2 , as x^2 is the only term. I wasn't integrating with a different variable.

  58. amistre64
    • 5 years ago
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    yay!! lol :) can I stop now lol

  59. anonymous
    • 5 years ago
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    ok

  60. anonymous
    • 5 years ago
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    wait that doesnt work though

  61. anonymous
    • 5 years ago
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    I mixed up a minus and plus sign ¬_¬

  62. anonymous
    • 5 years ago
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    its incorrect to do 1 / ( x^2 (x^2-15 ) = A / x^2 + b / (x^2-15) , thats incorrect

  63. anonymous
    • 5 years ago
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    you wont be able to find A and B

  64. anonymous
    • 5 years ago
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    I just did?

  65. anonymous
    • 5 years ago
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    you used a simpler case though

  66. anonymous
    • 5 years ago
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    Multiply it out, plz.

  67. anonymous
    • 5 years ago
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    1 = A(x^2 - 15) + B x^2

  68. anonymous
    • 5 years ago
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    , 1 = Ax^2 + bx^2 - 15

  69. anonymous
    • 5 years ago
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    (the + the last fraction should be minus, as in the first time, typo)

  70. anonymous
    • 5 years ago
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    1 = Ax^2 + Bx62 - 15 A

  71. anonymous
    • 5 years ago
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    hmmmm,

  72. anonymous
    • 5 years ago
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    but the book says it should be A/ x + B/x^2 + C x+d / (x^2 - 15)

  73. anonymous
    • 5 years ago
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    well this approach works as well

  74. anonymous
    • 5 years ago
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    You can do it if you want, but you'll find A = 0, so it doesn't really matter.

  75. anonymous
    • 5 years ago
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    right, so come the theorem doesnt work in these websites

  76. anonymous
    • 5 years ago
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    as far as repeated roots go

  77. anonymous
    • 5 years ago
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    for instance you didnt use CX+D, i think its because of that y = x^2 business,

  78. anonymous
    • 5 years ago
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    There are only terms in x^2 in this case, it wouldn't always work. The general method is probably more ... general.

  79. anonymous
    • 5 years ago
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    ahh

  80. anonymous
    • 5 years ago
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    so my question, my original question, hmmmm, do the quadratics have to be irreducible in the general case

  81. anonymous
    • 5 years ago
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    ok for instance, 1 / (( x-2)^2 ( x^2-15))

  82. anonymous
    • 5 years ago
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    1 / (( x-2)^2 ( x^2-15)) = A / (x-2) + b / (x-2)^2 + (cx+d) / (x^2 - 15 ) ?

  83. anonymous
    • 5 years ago
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    I'd say it's probably easier to leave it as the irreducible (without using difference of two squares) quadratic with Cx + d. I guess you could split up this factor, but it's unnecessary (as far as I know)

  84. anonymous
    • 5 years ago
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    Yes, that.

  85. anonymous
    • 5 years ago
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    but its not technically irreducible

  86. anonymous
    • 5 years ago
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    i can even go so far as to do this

  87. anonymous
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    1 / (( x-2)^2 ( x^2-15)) = (Ax+b) / (x-2)^2 + (cx+d) / (x^2 - 15 )

  88. anonymous
    • 5 years ago
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    no chat room

  89. anonymous
    • 5 years ago
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    sorry, not for you

  90. anonymous
    • 5 years ago
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    so this whole irreducible business is unnecessary

  91. anonymous
    • 5 years ago
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    1 = a*x^3-15*a*x+b*x^2-15*b+c*x^3-4*c*x^2+4*c*x+d*x^2-4*d*x+4*d

  92. anonymous
    • 5 years ago
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    ok im going to solve an easier one

  93. anonymous
    • 5 years ago
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    int 1/ ((x-2)(x^2-15))

  94. anonymous
    • 5 years ago
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    1/ ((x-2)(x^2-15)) = A/ (x-2) + (Bx+c) / ( x^ 2 - 15)

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