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anonymous

  • one year ago

Express following in partial fractions...evaluate the given integral

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  1. anonymous
    • one year ago
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    |dw:1438068450403:dw|

  2. anonymous
    • one year ago
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    @UsukiDoll

  3. UsukiDoll
    • one year ago
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    oh crap... let me expand (x-1)^2(x+1) first with wolfram and see if it comes back as factorable all the way.

  4. UsukiDoll
    • one year ago
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    ok it's all factorable so we have single letters to deal with.. one problem whenever we have a repeater... we have to take the original and build up the exponents.

  5. UsukiDoll
    • one year ago
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    for example |dw:1438068733033:dw| then take lcd of all of these guys

  6. UsukiDoll
    • one year ago
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    luckily the problem we are given isn't harsh like that (though I did s^5 at one point)

  7. UsukiDoll
    • one year ago
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    show me your attempt :)

  8. anonymous
    • one year ago
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    ok hold on

  9. anonymous
    • one year ago
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    \[\frac{ 2x }{ (x-2)^2(x+1) }=\frac{ A }{ x+1 }+\frac{ B }{(x-2)^2 }\] simplify \[2x=\frac{ A(x+2 )((x-2)^2) }{ x+2 } + \frac{ B((x-2)^2(x+2)) }{ (x-2)^2 }\]

  10. anonymous
    • one year ago
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    and then

  11. UsukiDoll
    • one year ago
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    STOP!

  12. UsukiDoll
    • one year ago
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    the A part is fine.. the one with the exponent.. no...

  13. anonymous
    • one year ago
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    \[2x=A(x-2)^2+B(x+1)\] x=2 gives \[4=A(0)^2+B(3)\] B=4/3

  14. anonymous
    • one year ago
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    ohh

  15. UsukiDoll
    • one year ago
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    whenever you have the exponent .. you have to write the single version first and then the exponent (depending on the number) look at this example |dw:1438069380011:dw|

  16. UsukiDoll
    • one year ago
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    I wrote the single version s and then work the exponent up s s^2 s^3 s^4 then you got to take lcd of all those fractions

  17. UsukiDoll
    • one year ago
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    |dw:1438069450631:dw|

  18. UsukiDoll
    • one year ago
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    I think order matters too... |dw:1438069483088:dw|

  19. UsukiDoll
    • one year ago
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    the lcd for this is \[(x+1)(x-1)^2 \] alright take the lcd ;)

  20. UsukiDoll
    • one year ago
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    ahem... take the lcd for |dw:1438069618368:dw|

  21. anonymous
    • one year ago
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    alright alright, give me a sec:)

  22. anonymous
    • one year ago
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    |dw:1438069792243:dw|

  23. UsukiDoll
    • one year ago
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    stop setting 2x = that's not going to work

  24. UsukiDoll
    • one year ago
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    the lcd part is good though

  25. UsukiDoll
    • one year ago
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    |dw:1438069902990:dw|

  26. UsukiDoll
    • one year ago
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    k expand all of that

  27. UsukiDoll
    • one year ago
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    *eats popcorn while I wait*

  28. anonymous
    • one year ago
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    I dont get this at all....whats wrong with me!!

  29. UsukiDoll
    • one year ago
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    oyyyyyyyy........

  30. UsukiDoll
    • one year ago
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    |dw:1438070102575:dw|

  31. UsukiDoll
    • one year ago
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    your turn :)

  32. anonymous
    • one year ago
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    whats does foil mean':)?

  33. UsukiDoll
    • one year ago
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    foil - first outer inner last... okkk expand :)

  34. anonymous
    • one year ago
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    Maybe I shouldve told u havent slept in 2 days...

  35. UsukiDoll
    • one year ago
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    then you should take a break from math and sleep

  36. anonymous
    • one year ago
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    its 10am...my exam is in 1 month from today....I have alot to do...due to circumstances...I cant sleep til 12hours from now,

  37. UsukiDoll
    • one year ago
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    try finishing the problem?

  38. anonymous
    • one year ago
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    so expanding c gives x^2-2x+1

  39. anonymous
    • one year ago
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    for A gives x^2-1

  40. UsukiDoll
    • one year ago
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    |dw:1438070573305:dw| |dw:1438070618427:dw|

  41. UsukiDoll
    • one year ago
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    yeah...

  42. UsukiDoll
    • one year ago
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    |dw:1438070677726:dw|

  43. UsukiDoll
    • one year ago
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    now distribute the letters

  44. anonymous
    • one year ago
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    Im really sry, can you finish it? I cant think properly, after this im going to eat and take a break and then start this again fresh.

  45. UsukiDoll
    • one year ago
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    the problem is that I can't give direct answers... so can't finish it without getting busted.

  46. anonymous
    • one year ago
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    @aric200

  47. UsukiDoll
    • one year ago
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    @Xlegalize can you come back over here so we can finish this please?

  48. anonymous
    • one year ago
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    im here

  49. UsukiDoll
    • one year ago
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    ok so the last step we left off is |dw:1438073386251:dw| so what we need to do next is to distribute the letters

  50. UsukiDoll
    • one year ago
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    I'll give an example \[A(x^2-1) = Ax^2-A \]

  51. UsukiDoll
    • one year ago
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    then similarly for B and C \[B(x+1) \rightarrow Bx+B \] \[C(x^2-2x+1) \rightarrow Cx^2-C2x+C \]

  52. UsukiDoll
    • one year ago
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    yeahhh that 2xC combination looks gross... and awkward XD anyway now we can put that long string in the numerator .

  53. UsukiDoll
    • one year ago
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    \[\large \frac{Ax^2-A+Bx+B+Cx^2-2xC+C}{(x+1)(x-1)^2}\]

  54. UsukiDoll
    • one year ago
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    |dw:1438073620766:dw| on the original equation, we have a 2x so that would mean that all equations with x would be equal to 2. whereas the plain letters and the letters with x^2 will be set equal to 0

  55. UsukiDoll
    • one year ago
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    \[-A+B+C =0\] \[B-2C=2 \] \[A+C =0 \]

  56. UsukiDoll
    • one year ago
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    so now we have these systems of equations and we need to solve for A,B,C once we get A B C we can plug it back into the equation from earlier

  57. anonymous
    • one year ago
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    B=2+2c A=2+2c+c A+c=0 2+2c+c+c=0

  58. UsukiDoll
    • one year ago
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    let's try adding the first and third equation up \[-A+B+C =0 \] \[A+C =0\] the A cancels out and we are left with B+2C= 0

  59. anonymous
    • one year ago
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    ok that is easier.

  60. UsukiDoll
    • one year ago
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    so now adding B-2C=2 B+2C=0 we have 2B=2 B = 1

  61. UsukiDoll
    • one year ago
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    so now that B = 1 the only place I can put that is the second equation if I put this in the first equation I'm stuck.. because I don't know what C or A is...yet

  62. UsukiDoll
    • one year ago
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    B-2C=2 1-2C=2 |dw:1438074079535:dw|

  63. UsukiDoll
    • one year ago
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    so B = 1, C = -1/2 now we can get our A picking out an easier equation with A is the best method to getting what A is

  64. anonymous
    • one year ago
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    a=1/2

  65. UsukiDoll
    • one year ago
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    |dw:1438074167001:dw|

  66. UsukiDoll
    • one year ago
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    that's right.. now that A = 1/2, B = 1, and C = -1/2 we need to check if they satisfy the three equations. If they do then we can go back, plug and chug into the earlier equation, and use integration

  67. UsukiDoll
    • one year ago
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    \[A+C =0 \] since A = 1/2 and C = -1/2 \[\frac{1}{2}-\frac{1}{2}=0 \] \[0=0 \] \[-A+B+C =0 \] \[-\frac{1}{2}+1-\frac{1}{2} =0\] \[-\frac{1}{2}-\frac{1}{2}+1 =0\] \[-\frac{1}{2}-\frac{1}{2}+1 =0\] \[-\frac{2}{2}+1 =0\] \[-1+1 =0\] \[0 =0\] \[B-2C=2 \] B = 1 \[1-2\frac{-1}{2}=2 \] \[1+\frac{2}{2}=2 \] \[1+1=2 \] \[2=2 \] ok after that long verification latex string we can plug our values into the earlier equation and integrate

  68. UsukiDoll
    • one year ago
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    |dw:1438074628205:dw|

  69. UsukiDoll
    • one year ago
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    so after all that lcd solving for a b c we can finally integrate... YAY! 2 of them are logs.. one is u substitution

  70. UsukiDoll
    • one year ago
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    |dw:1438074757028:dw| so the red portion is the \[\frac{1}{(x-1)^2}\] part. we have to do a bit more work on it

  71. anonymous
    • one year ago
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    Go on:)

  72. anonymous
    • one year ago
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    yeah I also got those parts.

  73. anonymous
    • one year ago
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    the last one takes more work.

  74. UsukiDoll
    • one year ago
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    yeah x.x now.. using u sub. u = x-1 du = 1 dx

  75. UsukiDoll
    • one year ago
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    |dw:1438074892358:dw| I wanna recheck that just to be sure

  76. anonymous
    • one year ago
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    yep thats correct.

  77. UsukiDoll
    • one year ago
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    thanks :) so final answer after all that jazz is |dw:1438075034959:dw|

  78. anonymous
    • one year ago
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    I will take a break and go through this all again. So Im 100 on this THANKS!!!!

  79. UsukiDoll
    • one year ago
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    you're welcome :)

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