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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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    why don't you type in math. will help in viewing easily.

  2. anonymous
    • 5 years ago
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    the second one is the right way. How do you hope getting radicals will help the partial fraction??(Indeed, we use partial fractions for simplification!!!)

  3. anonymous
    • 5 years ago
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    hello

  4. anonymous
    • 5 years ago
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    whats wrong with partial fractions with radicals?

  5. anonymous
    • 5 years ago
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    i dont see any problem with the first way

  6. anonymous
    • 5 years ago
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    no problems but we avoid it as calculations get messy.

  7. anonymous
    • 5 years ago
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    my computer does the partial fractions integral 1/(x^2 (x^2-15)) = -1/(15x^2) + 1/(15(x^2-15))

  8. anonymous
    • 5 years ago
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    yah...that "seems" easy :) you can always take x^2=y during partial fracs.. then u get 1/(y(y-15)) which can be easily split into fracs.

  9. anonymous
    • 5 years ago
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    that doesnt disobey partial fractions rule

  10. anonymous
    • 5 years ago
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    i thought it was , it had to be linear factors and irreducibles

  11. anonymous
    • 5 years ago
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    like A/x + B/x^2 + C / (x-sqrt 15) + D/(x+sqrt 15)

  12. anonymous
    • 5 years ago
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    See...there are no so-called rules...at least not for me :) if we take radicals the math is correct but its unconventional

  13. anonymous
    • 5 years ago
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    cx+d is supposed to be only for irreducibles, i thought

  14. anonymous
    • 5 years ago
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    like integral 1/x(x^2+1) = A/x + bx+c / x^2 + 1

  15. anonymous
    • 5 years ago
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    so youre saying 1/x(x^2-1) = A/x + (bx+c) /( x^2 - 1)

  16. anonymous
    • 5 years ago
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    not only for irreducibles,

  17. anonymous
    • 5 years ago
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    im pretty sure thats what the theorem states

  18. anonymous
    • 5 years ago
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    maybe the theorem u saw said that we cx+d for irreducibles but it did not say that we cant use cx+d for non-irreducibles!!

  19. anonymous
    • 5 years ago
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    we use cx+d for irreducibles but we can also use it like (bx+c)/(x^2-1). Here the math is correct. But splitting like A/(x+1)+B/(x-1) is more advantageous, as it gives a good form whose integral is known.

  20. anonymous
    • 5 years ago
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    for example, this partial fractions fails 1/(x^2 (x-5) ) = A/x^2 + B/(x-5) , this fails

  21. anonymous
    • 5 years ago
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    oh because of the A/x + B/x^2

  22. anonymous
    • 5 years ago
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    yes

  23. anonymous
    • 5 years ago
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    hmmm

  24. anonymous
    • 5 years ago
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    ok so 1/[(x-a)^n (x^2 +bx +c)^n] = k1/(x-a) + k2/(x-a)^2 + ... kn/(x-a)^n + m1x+p1/(x^2 +bx +c) +...

  25. anonymous
    • 5 years ago
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    your upto kn is ok. but from m1,p1,...its getting complicated.

  26. anonymous
    • 5 years ago
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    yeah im running out of letters

  27. anonymous
    • 5 years ago
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    no. there will be cx+d term above and the denominator's power will increase from 1 to n

  28. 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... +

  29. 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... +(Bnx +Cn) / (x^2 +bx +c)^n

  30. anonymous
    • 5 years ago
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    your c1 and b1 terms are wrong

  31. anonymous
    • 5 years ago
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    why?

  32. anonymous
    • 5 years ago
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    there won't be any b1x ... also instead of c1 write it as b1x+c1

  33. 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... +(Bnx +Cn) / (x^2 +bx +c)^n

  34. anonymous
    • 5 years ago
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    yups. thats correct....:)

  35. anonymous
    • 5 years ago
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    so thats fine?

  36. anonymous
    • 5 years ago
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    oh i left out the parenthesees

  37. anonymous
    • 5 years ago
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    ok so thats fine for partial fractions , and x^2 +bx + c doesnt necessarily have to be irreducible,

  38. anonymous
    • 5 years ago
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    see..make sure that you know that WE USE cx+d in numerator FOR IRREDUCIBLES (in other words polynomials have only complex roots). WE COULD use cx+d also above REDUCIBLES since it is mathematically correct but don't use it as one of the partial fractions is not in its lowest degrees (i.e. cx+d over a reducible is NOT recommended) But if such a use simplifies a problem at hand then why not use it? Understand me?

  39. anonymous
    • 5 years ago
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    a reducible quadratic you mean

  40. anonymous
    • 5 years ago
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    reducible into linear factors

  41. anonymous
    • 5 years ago
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    yups yups a QUADRATIC can only have cx+d above it. yes reducible into linear REAL factors. Nobody wants to have complex factors here!

  42. anonymous
    • 5 years ago
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    ok, so then we could do 1/(x^2 (x^2 - 15)) , we could reduce it

  43. anonymous
    • 5 years ago
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    im doing 1 / ( x ( x^2 + 2x + 1) ) the latter is reducible, but im not going to . = AX+b / ( x^2 + 2x + 1) + C / x

  44. anonymous
    • 5 years ago
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    Mathematically permissible. But not recommended.

  45. anonymous
    • 5 years ago
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    ok lets try this one 1/ ( x (x^2+2x+1)) integrate that using parfrac

  46. anonymous
    • 5 years ago
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    int 1/ ( x (x^2+2x+1))= (AX+b) / ( x^2 + 2x + 1) + C / x

  47. anonymous
    • 5 years ago
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    in part fracs.. its (1/x)+((-x-2)/(x^2+2x+1)) the first term gives log|x| and the second make substitution x^2+2x=y

  48. anonymous
    • 5 years ago
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    right, or we could have factored x^2 + 2x + 1

  49. anonymous
    • 5 years ago
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    yes...

  50. anonymous
    • 5 years ago
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    wait thats wrong

  51. anonymous
    • 5 years ago
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    whats that substitution ?

  52. anonymous
    • 5 years ago
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    yupss....the subs wrong...

  53. anonymous
    • 5 years ago
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    we can do it however

  54. anonymous
    • 5 years ago
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    oh by the way, partial fraction doesnt help us with say 1/ (x^(3/2) (x-15))

  55. anonymous
    • 5 years ago
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    no.. since the denominator is not a POLYNOMIAL

  56. anonymous
    • 5 years ago
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    , ahh

  57. anonymous
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    both numer. and denom. must be polynomials

  58. anonymous
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    wait is that a precondition?

  59. anonymous
    • 5 years ago
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    you could have 1/(x*(x^2-sqrt 3)) for example

  60. anonymous
    • 5 years ago
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    YES partial fractions can help ONLY TO RATIONAL FUNCTIONS i.e. both numeratoer and denominator polynomials with denominator not being the zero polynomial

  61. anonymous
    • 5 years ago
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    ok

  62. anonymous
    • 5 years ago
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    just clearing that up

  63. anonymous
    • 5 years ago
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    by the way what's your age?

  64. anonymous
    • 5 years ago
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    25

  65. anonymous
    • 5 years ago
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    why do i sound dumb

  66. anonymous
    • 5 years ago
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    hahha..joking? i hope u understood.. ask for doubts...

  67. anonymous
    • 5 years ago
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    i am 17

  68. anonymous
    • 5 years ago
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    oh

  69. anonymous
    • 5 years ago
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    jeez, i thought you were older. you are wise beyond your years :)

  70. anonymous
    • 5 years ago
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    are you a mathematician? or math-related worker? u named urself cantorset...

  71. anonymous
    • 5 years ago
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    yes im a math tutor

  72. anonymous
    • 5 years ago
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    for a living, i was stumped because the student didnt want to reduce a quadratic

  73. anonymous
    • 5 years ago
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    and i thought, hey the theorem says so and so

  74. anonymous
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    it was reducible into irrational factors

  75. anonymous
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    the question u asked me? - yes it can be solved both ways

  76. anonymous
    • 5 years ago
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    see this problem we just did its a lot easier if we reduce the quadratic

  77. anonymous
    • 5 years ago
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    integral (1/(x(x^2+2x+1))

  78. anonymous
    • 5 years ago
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    yes its lot easier if we reduce quad... but i argued just because i wanted to show that cx+d hold good for reducible quadratics also..

  79. anonymous
    • 5 years ago
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    right so i guess it depends on the problem

  80. anonymous
    • 5 years ago
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    well its good to know the power of this theorem

  81. anonymous
    • 5 years ago
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    which area of math are you specialized in?

  82. anonymous
    • 5 years ago
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    i tutor mostly calculus ,

  83. anonymous
    • 5 years ago
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    wow..thats awesome..i love calculus..but know little..struggling still with single variable calculus

  84. anonymous
    • 5 years ago
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    youre amazing :)

  85. anonymous
    • 5 years ago
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    well if you need help, give me a buzz

  86. anonymous
    • 5 years ago
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    ok..thank you.

  87. anonymous
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    so yeah, this one is tougher to integrate , without reducing the quadratic

  88. anonymous
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    integral (1/x)+((-x-2)/(x^2+2x+1))

  89. anonymous
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    yups.

  90. anonymous
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    so the original question integral 1/(x^2 (x^2-15))

  91. anonymous
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    then reducing the quadratic is worse, lol

  92. anonymous
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    x^2=y, integr=(1/15)log|(y-15)/y|+constant

  93. anonymous
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    then you need 2x dx = dy

  94. anonymous
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    oh its not so simple ,

  95. anonymous
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    i.e. integral=\[\frac{1}{15}\ln \left| \frac{x^2-15}{x^2} \right|+constant\]

  96. anonymous
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    when you make a substitution, you have to do dy = 2x dx

  97. anonymous
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    SORRY!! out of my mind......u r right..

  98. anonymous
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    then we have another problem

  99. anonymous
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    we want dx by itself

  100. anonymous
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    dy/(2x) = dx

  101. anonymous
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    so we have dx = dy / (2sqrt y))

  102. anonymous
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    ok, lets not do the substitution then

  103. anonymous
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    \[\frac{1}{15x}+\frac{1}{30\sqrt{15}}\ln \left| \frac{x-\sqrt{15}}{x+\sqrt{15}} \right|+constant\]

  104. anonymous
    • 5 years ago
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    i think this is the answer.

  105. anonymous
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    let me check

  106. anonymous
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    i use maple

  107. anonymous
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    1/(15*x)-(1/225)*sqrt(15)*arctanh((1/15)*x*sqrt(15))

  108. anonymous
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    must be the same..

  109. anonymous
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    oh i messed up

  110. anonymous
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    so what did you use ?

  111. anonymous
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    \[\int\limits_{}^{}\frac{dx}{x^2-15}=\frac{1}{\sqrt{15}} \ln \left| \frac{x-\sqrt{15}}{x+\sqrt{15}} \right| +C\] i used this identity

  112. anonymous
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    but theres an x^2 in front

  113. anonymous
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    that is easy!!

  114. anonymous
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    \[\int\limits_{?}^{?}1/(x^2(x^2-15))\]

  115. anonymous
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    how do you do fraction in latex?

  116. anonymous
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    frac{2}{3}=2/3

  117. anonymous
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    1/(x^2(x^2-15)) = A/x + B/x^2 + (cx+d) / (x^2-15)

  118. anonymous
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    interestingly enough, A = 0

  119. anonymous
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    so its equal to integra (-1/(15x) + (1/(15(x^2-15))

  120. anonymous
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    x^2=t substitution would save time. By the way, i hope the integrations are clear... i hav to leave...

  121. anonymous
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    but you have to do a little work with the substitution

  122. anonymous
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    yes..either way its correct.

  123. anonymous
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    oh i think you got it

  124. anonymous
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    so you fixed the substitution?

  125. anonymous
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    yes..

  126. anonymous
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    but thats weird

  127. anonymous
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    ok x^2 = y, so dx = dy/(2x) = dy/(2sqrt y )

  128. anonymous
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    so int 1/(x^2(x^2-15)) = int (1/( 2 sqrt y *y *(y-15))

  129. anonymous
    • 5 years ago
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    i used the substitution only for the partial fractions. not for the integrations. i am working with y in partial fracs but wreturning to x in integration

  130. anonymous
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    oh

  131. anonymous
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    it saves time in partial fracs..

  132. anonymous
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    so you got integral 1/(x^2(x^2-15)) = integral (-1/(15x) + (1/(15(x^2-15)). and THEN did substitution

  133. anonymous
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    no ..i used y=x^2 only for partial fracs..

  134. anonymous
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    you cant do that, as far as i know

  135. anonymous
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    That is allowed.

  136. anonymous
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    oh then you substituted back before you integrated?

  137. anonymous
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    working with a smaller degree variable in partial frac.. is very good idea. yah i subst. back before i integrated

  138. anonymous
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    ok so you got integral 1/(x^2(x^2-15)) = integral (-1/(15y) + (1/(15(y-15)).

  139. anonymous
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    Yes. that's right. i then changed y to x^2

  140. anonymous
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    thats the same thing as just integral 1/(x^2(x^2-15)) = integral (-1/(15x^2) + (1/(15(x^2-15))

  141. anonymous
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    oh i see, since its similiar to 1 / ( x (x-5))

  142. anonymous
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    yes...but during partial fracs.. dealing with y is easier than dealing with x^2...its just the smae thing

  143. anonymous
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    you are right

  144. anonymous
    • 5 years ago
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    right, but one has to be careful to switch back

  145. anonymous
    • 5 years ago
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    yes...i did a silly mistake back 15 mins. earlier

  146. anonymous
    • 5 years ago
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    youre basically using the identity 1/(u(u-15) = -1/15u + 1/15(u-15)

  147. anonymous
    • 5 years ago
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    yes

  148. anonymous
    • 5 years ago
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    ok

  149. anonymous
    • 5 years ago
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    substitution is way for complicated, i mean u substitution

  150. anonymous
    • 5 years ago
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    yes, but helps sometimes...avoid it when unnecessary...

  151. anonymous
    • 5 years ago
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    right, if we did u substitution, we would be able to do partial fractions anymore

  152. anonymous
    • 5 years ago
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    we would have u = x^2 , du = 2x dx, and dx = du/(2x) = du / (2 sqrt u) int ( 1/ (sqrt u * u * (u-15)) du

  153. anonymous
    • 5 years ago
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    yes..

  154. anonymous
    • 5 years ago
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    this is a very interesting problem, the pitfalls

  155. anonymous
    • 5 years ago
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    pitfall is like quicksand, lol

  156. anonymous
    • 5 years ago
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    so its better to say, i used the algebraic identity, blah blah

  157. anonymous
    • 5 years ago
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    one has to be lucky to get the right idea for integration

  158. anonymous
    • 5 years ago
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    so you dont confuse , lol

  159. anonymous
    • 5 years ago
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    i substituted into the algebraic identity, that would be best :)

  160. anonymous
    • 5 years ago
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    1/(u*(u-15)) = 1/ (15 u ) + ./.. etc

  161. anonymous
    • 5 years ago
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    whatever...everyone has a different way to approach problems ..:)

  162. anonymous
    • 5 years ago
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    thats fine

  163. anonymous
    • 5 years ago
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    its just when you say substitution and integration, most people would think you are doing u substitution

  164. anonymous
    • 5 years ago
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    really what it should be called is change of variable

  165. anonymous
    • 5 years ago
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    anyways, i see now why partial fractions fail on the complete u substitution

  166. anonymous
    • 5 years ago
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    its not a rational function, as you pointed out. thanks for your help. you've been great!

  167. anonymous
    • 5 years ago
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    yah...i should have shown the whole process but takes too much time in latex..then there would have been no confusions...by the way i m not too good in english

  168. anonymous
    • 5 years ago
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    english aint bad , lol

  169. anonymous
    • 5 years ago
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    what country?

  170. anonymous
    • 5 years ago
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    where are u from? i am from india

  171. anonymous
    • 5 years ago
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    USA

  172. anonymous
    • 5 years ago
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    near new york

  173. anonymous
    • 5 years ago
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    k...celebrations there right? after osama was killed?

  174. anonymous
    • 5 years ago
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    i didnt celebrate

  175. anonymous
    • 5 years ago
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    he wasnt given a trial, that kind of thing

  176. anonymous
    • 5 years ago
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    i would like to see the evidence, and charges. but i guess it was stupid for him to take credit for it, if he really did take credit

  177. anonymous
    • 5 years ago
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    he could've escaped if he was allowed to live....nobody can predict him and his associates,....:)

  178. anonymous
    • 5 years ago
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    but he was a CIA asset, when he was fighting the russians

  179. anonymous
    • 5 years ago
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    im not crazy about our govt. too much deception

  180. anonymous
    • 5 years ago
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    i dont know if he is solely responsible for 911 , he might just be one piece of the puzzle

  181. anonymous
    • 5 years ago
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    whatever lets leave these to the concerned authorities... Hope u r ok with par fracs...bye

  182. anonymous
    • 5 years ago
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    take care

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spraguer (Moderator)
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