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Christos Group Title

Please fellow open study members, solve with me . Finals countdown 3 hours :P https://www.dropbox.com/s/vwu945w28tgowlp/Screenshot%202014-01-15%2015.09.53.jpg (ANXIETY LEVEL OVER THE TOP)

  • 10 months ago
  • 10 months ago

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  1. hartnn Group Title
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    so where are you stuck ?

    • 10 months ago
  2. Christos Group Title
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    @hartnn

    • 10 months ago
  3. Christos Group Title
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    They are the only problems to which I don't have solutions , and I also have infinite series to cover , if you give me what I lack, it would be of a great help at the current situation ! :(

    • 10 months ago
  4. ganeshie8 Group Title
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    link is not opening for meh

    • 10 months ago
  5. Christos Group Title
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    http://screencast.com/t/WittgfBhztpV

    • 10 months ago
  6. Christos Group Title
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    just tell me what do I do

    • 10 months ago
  7. Christos Group Title
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    for the third one

    • 10 months ago
  8. Christos Group Title
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    after the long division , then what

    • 10 months ago
  9. ganeshie8 Group Title
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    before we dive in, two things to keep in mind :- 1) we can integrate ANY/EVERY rational function 2) to use partial fractions, degreem of numerator must always be LESS than degree of denominator

    • 10 months ago
  10. ganeshie8 Group Title
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    *degree

    • 10 months ago
  11. Christos Group Title
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    *reading carefully*

    • 10 months ago
  12. hartnn Group Title
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    yeah, thats the reason she did long division for 3rd

    • 10 months ago
  13. hartnn Group Title
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    so, you must have something like x/x^2+1 ,right ??

    • 10 months ago
  14. hartnn Group Title
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    hint : d/dx (x^2+1) = 2x

    • 10 months ago
  15. hartnn Group Title
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    so take u = x^2+1 , du =...

    • 10 months ago
  16. ganeshie8 Group Title
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    for first example, since degree of numerator < degree of denominator, next do below steps :- step1 : factor the denominator ive seen hartnn explaining partial fractions excellently before... so i give u hartnn :)

    • 10 months ago
  17. ganeshie8 Group Title
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    @hartnn ... plz continue if u have time :)

    • 10 months ago
  18. hartnn Group Title
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    i have like 3 minutes :P

    • 10 months ago
  19. ganeshie8 Group Title
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    and im sure, u can make Christos understand half of calculus in less than 3 minutes ;)

    • 10 months ago
  20. hartnn Group Title
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    probably, if she is sitting right in front of me ;)

    • 10 months ago
  21. Christos Group Title
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    she is intenting a new study method called parallel processing !!

    • 10 months ago
  22. Christos Group Title
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    I am just kidding

    • 10 months ago
  23. hartnn Group Title
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    times up :P i need to rush b4 my bus leaves without me :P be back online when i reach home..

    • 10 months ago
  24. ganeshie8 Group Title
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    oh hartnn you're in office ha ... runn.. il see if i can do some justice in explaining partial fractions :)

    • 10 months ago
  25. ganeshie8 Group Title
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    \(\large \frac{1}{x^2+3x-4}\) factor the denominator \(\large \frac{1}{(x-1)(x+4)}\)

    • 10 months ago
  26. ganeshie8 Group Title
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    step2 : break left fraction into sum of fractions :- \(\large \frac{1}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4}\) find A and B

    • 10 months ago
  27. Christos Group Title
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    answer: 1/5ln|x-2| - 1/5 ln|x+4| + C

    • 10 months ago
  28. Christos Group Title
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    is it ?

    • 10 months ago
  29. ganeshie8 Group Title
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    thats right !

    • 10 months ago
  30. Christos Group Title
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    lets go 2nd

    • 10 months ago
  31. Christos Group Title
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    do I need to factor the denominator here ?

    • 10 months ago
  32. ganeshie8 Group Title
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    yes, factoring IS the first step

    • 10 months ago
  33. ganeshie8 Group Title
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    and ofc, the 0th step is long-division :)

    • 10 months ago
  34. Christos Group Title
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    A/x b/x+1 C/x+1

    • 10 months ago
  35. Christos Group Title
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    B/x-1 * * *

    • 10 months ago
  36. ganeshie8 Group Title
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    step1 : factor denominator \(\large \frac{1}{x(x^2-1)} \) \(\large \frac{1}{x(x-1)(x+1)} \)

    • 10 months ago
  37. Christos Group Title
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    A B C: http://screencast.com/t/a5HayvjoOi

    • 10 months ago
  38. ganeshie8 Group Title
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    step2 : wrote the fraction as sum of simple fractions \(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) solve A, B, C

    • 10 months ago
  39. ganeshie8 Group Title
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    yes looks good, solve A, B, C

    • 10 months ago
  40. Christos Group Title
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    I hope you are not leading me into a teaching trap at the current moment huh :D

    • 10 months ago
  41. Christos Group Title
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    ok ok I solve it now

    • 10 months ago
  42. ganeshie8 Group Title
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    whats teaching trap lol im no teacher... guess im more toward a professor who screws his students hmm lol

    • 10 months ago
  43. Christos Group Title
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    ok I can't solve this, whats the next step :D

    • 10 months ago
  44. Christos Group Title
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    if x = 1 denominator is undefined A.k.A me missing something

    • 10 months ago
  45. ganeshie8 Group Title
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    there is a small trick to solve this, it wont take any time to learn the trick. let me show u how to find A, B, C super fast

    • 10 months ago
  46. ganeshie8 Group Title
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    \(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) to sovle A : Multuply the whole equation wid \(x\), and plug \(x = 0\) \(\large \frac{1}{(x-1)(x+1)} = A + \frac{Bx}{x-1} + \frac{Cx}{x+1}\)

    • 10 months ago
  47. ganeshie8 Group Title
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    when u plug \(x = 0\), everything on right side cancels away, except \(A\)

    • 10 months ago
  48. Christos Group Title
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    -1

    • 10 months ago
  49. ganeshie8 Group Title
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    \(\large \frac{1}{(0-1)(0+1)} = A + 0 + 0\) solve A

    • 10 months ago
  50. ganeshie8 Group Title
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    yes, A = -1

    • 10 months ago
  51. Christos Group Title
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    yesyes

    • 10 months ago
  52. ganeshie8 Group Title
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    similarly find B and C

    • 10 months ago
  53. Christos Group Title
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    x = ?

    • 10 months ago
  54. ganeshie8 Group Title
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    \(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) to find B :- multiply the whole equation wid \((x-1)\) and plugin \(x = 1\) \(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\)

    • 10 months ago
  55. Christos Group Title
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    and then the main fraction is undefined..

    • 10 months ago
  56. ganeshie8 Group Title
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    how ?

    • 10 months ago
  57. ganeshie8 Group Title
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    \(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\) plugin x = 1 \(\large \frac{1}{1(1+1)} = 0 + B + 0\)

    • 10 months ago
  58. ganeshie8 Group Title
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    il let u solve C :)

    • 10 months ago
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