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anonymous

  • one year ago

By expressing the following in partial fractions evaluate the given intergral.

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  1. anonymous
    • one year ago
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    |dw:1437741496765:dw|

  2. UsukiDoll
    • one year ago
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    so first we have to take partial fractions before we can take the antiderivative. we can ignore the integral symbol for the first half of this problem

  3. UsukiDoll
    • one year ago
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    now there are two forms of partial fractions: factored and non-factored we have to look closely in the denominator.. if we can factor something out. our numerator will just be a single letter like A for example suppose we are given this fraction \[\frac{1}{x^2-1}\] we can factor \[x^2-1 = (x+1)(x-1) \] \[\frac{1}{(x+1)(x-1)}\] so for each partial fraction we have \[\frac{A}{x+1} +\frac{B}{x-1}\]

  4. UsukiDoll
    • one year ago
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    now we have to take the LCD which is (x-1)(x+1) \[\frac{x-1}{x-1} \cdot \frac{A}{x+1} +\frac{B}{x-1} \cdot ]\frac{x+1}{x+1}\]

  5. anonymous
    • one year ago
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    I dont understand how u got from 1/(x^3+x) to 1/(x^2-1) the rest u wrote below, Im familiar with.

  6. UsukiDoll
    • one year ago
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    I am providing an example first anyway then we distribute the A and B \[\frac{Ax-A+Bx+B}{(x-1)(x+1)}\]

  7. UsukiDoll
    • one year ago
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    so we have these equations -A+B = 1 A+B = 0 we got the one from the numerator of the original equation

  8. anonymous
    • one year ago
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    ah ok that makes sense.

  9. UsukiDoll
    • one year ago
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    so now we solve these system of equations through elimination method. That would give us 2B=1 or B =1/2 so we plug that value in one of the equations \[A+\frac{1}{2}=0\] so our \[A = \frac{-1}{2}\]

  10. UsukiDoll
    • one year ago
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    yes I like to give a guideline first before we go to the actual problem

  11. anonymous
    • one year ago
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    that example makes actual sense, my book skipped alot of steps u took.

  12. UsukiDoll
    • one year ago
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    oh dear T_T glad I came lol so we plug in A = -1/2 and B = 1/2 back into the equation before we took the LCD \[\large \frac{\frac{-1}{2}}{x+1} +\frac{\frac{1}{2}}{x-1}\]

  13. UsukiDoll
    • one year ago
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    then we take the antiderivative which is just logs

  14. UsukiDoll
    • one year ago
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    \[\frac{1}{2} \ln(x+1)-\frac{1}{2}\ln(x-1) +C \]

  15. anonymous
    • one year ago
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    x+1 is under -1/2 dont u mean vice versa with the pos/negative signs on the 1/2s?

  16. UsukiDoll
    • one year ago
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    oh I missed the sign sorry

  17. UsukiDoll
    • one year ago
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    \[\frac{-1}{2} \ln(x+1)-\frac{1}{2}\ln(x-1) +C\]

  18. UsukiDoll
    • one year ago
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    so let's apply what we've done to \[\large \frac{1}{x^3+x}\]

  19. UsukiDoll
    • one year ago
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    can we factor the denominator?

  20. anonymous
    • one year ago
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    x(x^2+1)

  21. UsukiDoll
    • one year ago
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    woo hoo! that means we can use single letters ^_^ should it be non-factorable we have to use Ax+B :(

  22. UsukiDoll
    • one year ago
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    if I make typos, please stop me. It's almost 3 am and I am typing fast

  23. UsukiDoll
    • one year ago
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    so anyway

  24. anonymous
    • one year ago
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    Haha, okey :)

  25. UsukiDoll
    • one year ago
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    \[\large \frac{1}{x(x^2+1)}\]

  26. UsukiDoll
    • one year ago
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    so since it's factorable we use single letters \[\frac{A}{x}+\frac{B}{x^2+1}\]

  27. UsukiDoll
    • one year ago
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    so now we use LCD

  28. UsukiDoll
    • one year ago
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    \[\frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{B}{x^2+1} \cdot \frac{x}{x}\]

  29. anonymous
    • one year ago
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    cant I just skip that and write A(x^2+1)+bx

  30. UsukiDoll
    • one year ago
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    then we distribute \[\frac{Ax^2+A+Bx}{x(x^2+1)}\] I don't know if it's a good idea though. skipping steps at this topic is not recommended in my opinion

  31. anonymous
    • one year ago
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    okey, I will follow through then:)

  32. UsukiDoll
    • one year ago
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    hmmm ... the thing is we have these equations A = 1 Bx = 0 Ax^2 = 0

  33. IrishBoy123
    • one year ago
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    you have to use: \(\large \frac{A}{x}+\frac{Bx + C}{x^2+1}\)

  34. IrishBoy123
    • one year ago
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    you have a quadratic on the bottom of the second term so you need a linear term on top

  35. UsukiDoll
    • one year ago
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    ah geez... no problem... OH that's right x^2+1 doesn't factor thanks for the head up

  36. UsukiDoll
    • one year ago
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    yeah x^2+1 doesn't factor so Bx+C so \[x^3+x \rightarrow (x)(x^2+1) \] a portion is factored. we can take an x out but we can't do anything to x^2+1

  37. UsukiDoll
    • one year ago
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    \[\large \frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{Bx + C}{x^2+1} \cdot \frac{x}{x}\]

  38. UsukiDoll
    • one year ago
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    there we go now we can distribute \[\large \frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}\]

  39. UsukiDoll
    • one year ago
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    now ...... we just have one value from the original equation \[\large \frac{1}{x^3+x} \] So we have three equations A = 1 C =0 A+B = 0

  40. UsukiDoll
    • one year ago
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    you're probably wondering why 1? why is the rest of the equations = 0 it's because our numerator is just that number if we have something like 1 + 2x on the numerator the C = 2 due to the fact that we had Cx after we did partial fractions

  41. UsukiDoll
    • one year ago
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    It's like this A= 1 Cx = 0 Ax^2+Bx^2 = 0 because there is no number next to the x and no number next to the x^2 so they are set to 0... we don't have to write the variable of x's for these letter equations

  42. anonymous
    • one year ago
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    ohh okey

  43. UsukiDoll
    • one year ago
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    so we know that A = 1 C = 0 A+B = 0 we know C and A so we solve for B 1+B=0 B = -1

  44. UsukiDoll
    • one year ago
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    so we plug in A = 1, B =-1, and C = 0 back into the corrected equation shout out to @IrishBoy123 !!!!!!!! WOOOOOOOOOOOOOOOO

  45. UsukiDoll
    • one year ago
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    \[\large \frac{A}{x}+\frac{Bx + C}{x^2+1}\] \[\large \frac{-1}{x}+\frac{1x + 0}{x^2+1}\] \[\large \large \frac{-1}{x}+\frac{x}{x^2+1}\]

  46. UsukiDoll
    • one year ago
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    now we can take antiderivatives... finally after that long journey we CAN INTEGRATE!

  47. UsukiDoll
    • one year ago
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    first one is just ln x \[\large - \ln x+\frac{x}{x^2+1}\]

  48. UsukiDoll
    • one year ago
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    the second one is u-substitution

  49. IrishBoy123
    • one year ago
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    A = 1, B =-1

  50. UsukiDoll
    • one year ago
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    yeah I got that @IrishBoy123

  51. UsukiDoll
    • one year ago
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    A =1 , B =-1, and C = 0

  52. UsukiDoll
    • one year ago
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    but thanks for the verification :)

  53. anonymous
    • one year ago
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    \[\frac{ -1 }{ x^2 +1} is...?\]

  54. UsukiDoll
    • one year ago
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    so back to u-substitution \[\large u=x^2+1 \]

  55. UsukiDoll
    • one year ago
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    \[\large - \ln x+\frac{x}{x^2+1} \] u-subsitution \[\large u=x^2+1 \] \[\large du = 2x dx \] \[\large \frac{1}{2}du = x dx \] \[\large - \ln x+(\frac{1}{2})\frac{1}{u} \]

  56. UsukiDoll
    • one year ago
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    the antiderivative of 1/u is ln u

  57. UsukiDoll
    • one year ago
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    \[\large - \ln x+(\frac{1}{2}) (\ln u) +C\] then subsitute u back in which was \[\large u=x^2+1 \] \[\large - \ln x+(\frac{1}{2}) (\ln x^2+1) +C\]

  58. UsukiDoll
    • one year ago
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    whew

  59. UsukiDoll
    • one year ago
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    oops miss the l l \[\large - \ln \mid x \mid +(\frac{1}{2}) (\ln \mid x^2+1 \mid) +C\]

  60. UsukiDoll
    • one year ago
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    so to check we just take the derivative.. it should come out to be what we had before

  61. anonymous
    • one year ago
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    U spent over 30min for this question DAAMN. Thanks bro, Your a lifesaver.

  62. UsukiDoll
    • one year ago
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    \[\large -\frac{1}{x} + \frac{1}{2}\frac{1}{x^2+1}(2x) + \ln \mid x^2+1 \mid(0) \] then we got \[\large -\frac{1}{x}+\frac{x}{x^2+1}\]

  63. UsukiDoll
    • one year ago
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    which is what we had earlier... thanks XD

  64. anonymous
    • one year ago
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    yup, got the correct answer as the book asked for. R u a math teacher? or tuitor of any kind?

  65. UsukiDoll
    • one year ago
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    math major going into teaching.

  66. UsukiDoll
    • one year ago
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    Partial Fractions shows up again in Ordinary Differential Equations.

  67. anonymous
    • one year ago
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    I will think of you when they do :D

  68. anonymous
    • one year ago
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    Alright closing this, THANK YOU <3

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