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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. anonymous
    • 5 years ago
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    no

  2. anonymous
    • 5 years ago
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    either way worked out

  3. anonymous
    • 5 years ago
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    i checked my program

  4. anonymous
    • 5 years ago
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    \[1/15\int\limits_{}^{}(x ^{2}-(x ^{2}-15))dx/(x ^{2}(x ^{2}-15))\]\[1/15[\int\limits_{}^{}dx/(x ^{2}-15)-\int\limits_{}^{}dx/x ^{2}\]

  5. anonymous
    • 5 years ago
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    now go ahead.

  6. anonymous
    • 5 years ago
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    yes that works, but i want to use partial fractions

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

  8. anonymous
    • 5 years ago
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    thats some trick you got there :)

  9. anonymous
    • 5 years ago
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    partial fraction will be avoided when you have better choices...

  10. anonymous
    • 5 years ago
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    the tricks will help u and its very time saving...

  11. anonymous
    • 5 years ago
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    i guess was wondering, is it ok to have , a quadratic that is not reducible , ok let me state this correctly

  12. anonymous
    • 5 years ago
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    how did you know to seperate them like that

  13. anonymous
    • 5 years ago
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    only practice..........

  14. anonymous
    • 5 years ago
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    i can make it harder, one sec

  15. anonymous
    • 5 years ago
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    int (1/ (x^3 (x^2-15))

  16. anonymous
    • 5 years ago
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    ok thats a good partial fractions problem

  17. anonymous
    • 5 years ago
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    yea...

  18. anonymous
    • 5 years ago
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    here you can use partial fractions because no other way............

  19. anonymous
    • 5 years ago
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    ok, now we have two options it would seem, my question really boils down to this quadratic

  20. anonymous
    • 5 years ago
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    the quadratic is not irreducible,

  21. anonymous
    • 5 years ago
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    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^3 +D/(x-sqrt 15) + e/(x+sqrt 15) , or 1/((x^2*(x^2-15)) = A / x + B/x^2 + Cx+d / ( x^2 - 15)

  22. anonymous
    • 5 years ago
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    here u can use the one which i gave first to reduce it . it would give you \[1/15[\int\limits_{}^{}dx/(x(x ^{2}-15)-\int\limits_{}^{}dx/x ^{3}]\]

  23. anonymous
    • 5 years ago
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    my question was, both approaches are legal?

  24. anonymous
    • 5 years ago
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    does that work, im not sure

  25. anonymous
    • 5 years ago
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    both are correct.

  26. anonymous
    • 5 years ago
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    oh ok, i thought the quadratic MUST be irreducible

  27. anonymous
    • 5 years ago
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    but when u try to find cx+d/(x^2-15) then may be again u have to do partial fractions...

  28. anonymous
    • 5 years ago
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    are you sure this is correct 1/15[∫dx/(x(x2−15)−∫dx/x3], i did it and got a different answer

  29. anonymous
    • 5 years ago
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    what did u get?

  30. anonymous
    • 5 years ago
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    [ x^3 - x(x^2-15) ] / (x^4 (x^2 - 15)

  31. anonymous
    • 5 years ago
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    no...

  32. anonymous
    • 5 years ago
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    im trying to reverse your result, checking if its equivlanet

  33. anonymous
    • 5 years ago
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    ((x^2)-(x^2-15))/(x^3(x^2-15))

  34. anonymous
    • 5 years ago
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    oh it works

  35. anonymous
    • 5 years ago
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    ok

  36. anonymous
    • 5 years ago
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    my math was off,

  37. anonymous
    • 5 years ago
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    again there is another way to do it just substitute x^2=y.

  38. anonymous
    • 5 years ago
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    thats not going to work, since 2x dx = dy

  39. anonymous
    • 5 years ago
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    it is again more easier ....

  40. anonymous
    • 5 years ago
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    then dx = dy / (2x )

  41. anonymous
    • 5 years ago
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    oh you mean for the first problem ?

  42. anonymous
    • 5 years ago
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    your integration is xdx/(x^4(x^2-15)

  43. anonymous
    • 5 years ago
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    for the second one...

  44. anonymous
    • 5 years ago
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    try it... it surely works..

  45. anonymous
    • 5 years ago
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    oh , but its not simple

  46. anonymous
    • 5 years ago
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    lets try it for the first integral

  47. anonymous
    • 5 years ago
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    oh ok , so s

  48. anonymous
    • 5 years ago
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    its not for the first one.... its only applicable for the second one....

  49. anonymous
    • 5 years ago
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    these some shortcut to approach....

  50. anonymous
    • 5 years ago
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    i have 2 go now.....

  51. anonymous
    • 5 years ago
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    dx = dy / ( 2 sqrt y )

  52. anonymous
    • 5 years ago
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    do you agree to that?

  53. anonymous
    • 5 years ago
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    have u understood properly?

  54. anonymous
    • 5 years ago
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    yes, i just want to make sure i understand your subsitution

  55. anonymous
    • 5 years ago
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    dx/(x^3(x^2-15) gives you dy/(2y^2(y-15))

  56. anonymous
    • 5 years ago
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    i get dy / ( 2 y^(3/2) (y-15)

  57. anonymous
    • 5 years ago
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    y=x^2 dy=2xdx dx/(x^3(x^2-15))= 2xdx/(2x^4(x^2-15)) = dy/(2y^2(y-15))

  58. anonymous
    • 5 years ago
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    got it?

  59. anonymous
    • 5 years ago
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    thats wrong

  60. anonymous
    • 5 years ago
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    you have to solve for dx

  61. anonymous
    • 5 years ago
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    what?

  62. anonymous
    • 5 years ago
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    check it , thats correct...

  63. anonymous
    • 5 years ago
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    2x dx = dy , so dx = dy/(2x)

  64. anonymous
    • 5 years ago
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    ok...let me to make u clear...

  65. anonymous
    • 5 years ago
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    oh, you multiplied top and bottom by 2x, one sec

  66. anonymous
    • 5 years ago
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    yesssssssssss

  67. anonymous
    • 5 years ago
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    be fast i have to go .... but its nice to solve maths with you?

  68. anonymous
    • 5 years ago
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    very cleverl

  69. anonymous
    • 5 years ago
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    thanks , bye

  70. anonymous
    • 5 years ago
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    sure we will talk again

  71. anonymous
    • 5 years ago
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    thats the magic of maths...........

  72. anonymous
    • 5 years ago
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    my way probably works, i have to work on it :)

  73. anonymous
    • 5 years ago
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    my substitution

  74. anonymous
    • 5 years ago
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    what do u do?

  75. anonymous
    • 5 years ago
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    i solve for dx =

  76. anonymous
    • 5 years ago
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    from where r u?

  77. anonymous
    • 5 years ago
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    i live in usa, near new york

  78. anonymous
    • 5 years ago
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    in which class are u?

  79. anonymous
    • 5 years ago
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    2x dx = dy , so dx = dy/(2x) , so dx = dy/ ( 2 sqrt y)

  80. anonymous
    • 5 years ago
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    but it comes out the same

  81. anonymous
    • 5 years ago
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    since x^3 = sqrt y ^3

  82. anonymous
    • 5 years ago
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    etc

  83. anonymous
    • 5 years ago
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    are u a student in college or school?

  84. anonymous
    • 5 years ago
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    college

  85. anonymous
    • 5 years ago
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    yes thats right..

  86. anonymous
    • 5 years ago
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    very nice, both ways work

  87. anonymous
    • 5 years ago
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    ok bye.......

  88. anonymous
    • 5 years ago
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    bye

  89. anonymous
    • 5 years ago
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    i like your way though

  90. anonymous
    • 5 years ago
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    it never occured to me you can multiply top and bottom by x

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