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dessyj1

  • one year ago

Calculus 1. Question in the comments

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  1. dessyj1
    • one year ago
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    \[\int\limits_{none}^{none} \frac{ 11x+4 }{ (3+6x-x ^{2})^{2} } dx\]

  2. dessyj1
    • one year ago
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    this is supposed to be an indefinite integral.

  3. dessyj1
    • one year ago
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    I cannot easily do u-substitution

  4. anonymous
    • one year ago
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    Use partial fraction decomposition \[\frac{ 11x+4 }{(-x^2+6x+3)^2 }=\frac{ Ax+B }{ -x^x+6x+3 }+\frac{ Cx+D }{ (-x^2+6x+3)^2 }\]

  5. anonymous
    • one year ago
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    Multiply to eliminate fractions \[11x+4=(Ax+B)(-x^2+6x+3)+Cx+D\]

  6. dessyj1
    • one year ago
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    Would you mind showing that to me? I only learned u-substitution.

  7. dessyj1
    • one year ago
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    you mean -x^2 instand of x^x right?

  8. anonymous
    • one year ago
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    yes I do mean -x². Read up on it here: http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx There's a table in the middle of the page that shows what format the decomposed fraction is supposed to do. I'll work through this one.

  9. anonymous
    • one year ago
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    Once you have the fractions eliminated, multiply everything out \[11x+4=-Ax^3+6Ax^2+3Ax-Bx^2+6Bx+3B+Cx+D\] Group everything together by the x variable \[11x+4=x^3(-A)+x^2(6A-B)+x(3A+6B+C)+(3B+D)\] *check my multiplication please*

  10. dessyj1
    • one year ago
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    checks out.

  11. anonymous
    • one year ago
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    Now you can equate coefficients to create a system of equations. On the left, there is no x³ term and on the right the coefficient of x³ is -A, so the equation is \[0=-A\] For x², it's \[0=6A-B\] For x, \[11=3A+6B+C\] For the constant 4 = 3B+D

  12. dessyj1
    • one year ago
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    that makes sense

  13. anonymous
    • one year ago
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    When I solve the system I get A = 0, B = 0, C = 11, and D = 4, which puts us right where we started :/. PFD is usually a solid method for integrating rational functions. Let me see if I can figure another method

  14. dessyj1
    • one year ago
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    Did you solve the system of equations by going off of the fact that you knew A=0?

  15. anonymous
    • one year ago
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    yes I did. Ignore this picture for now |dw:1434688610771:dw|

  16. anonymous
    • one year ago
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    If we use u-substitution, \[u=-x^2+6x+3\] \[du=(-2x+6)dx\] We need to make the numerator 11x+4 look like -2x+6. Start by multiplying by 1 in the form of -2/-2 \[11(\frac{ -2 }{ -2}) x+4\]\[-\frac{ 11 }{ 2 }(-2x)+4\] I'm going to add 6 inside the parentheses, which is the same as adding -33 to the overall expression. I'll have to add 33 outside so the expression's value doesn't change \[-\frac{ 11 }{ 2 }(-2x+6)+37\]

  17. anonymous
    • one year ago
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    So now you're taking the integral of|dw:1434689368064:dw|

  18. anonymous
    • one year ago
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    you can use u-substitution for the integral on the left. And I'm going to check if I can find the one on the right in an integral table

  19. anonymous
    • one year ago
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    you with me so far?

  20. dessyj1
    • one year ago
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    I am.

  21. dessyj1
    • one year ago
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    wait.

  22. dessyj1
    • one year ago
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    why did you just multiply (-2/-2) to 11?

  23. anonymous
    • one year ago
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    because the coefficient of x in du is -2 and I needed it to match that.

  24. dessyj1
    • one year ago
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    why did it not apply to the 4 as well?

  25. anonymous
    • one year ago
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    It wouldn't have had any effect. The -2's would have just cancelled. Technically, they cancel when multiplied by 11, but I didn't because I want the coefficient to be -2.

  26. dessyj1
    • one year ago
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    did you try to balance that by adding 37?

  27. dessyj1
    • one year ago
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    if that is the case then i am with you.

  28. anonymous
    • one year ago
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    |dw:1434690426953:dw|

  29. anonymous
    • one year ago
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    yes, adding 37 balances it

  30. dessyj1
    • one year ago
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    forgot the neg 1 exp

  31. dessyj1
    • one year ago
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    and how did you know it would balance it?

  32. anonymous
    • one year ago
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    oh, sharp eye :) |dw:1434690646799:dw|

  33. anonymous
    • one year ago
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    because I added 6 inside the parentheses. Distributing -11/2 means I added -33 to the whole thing. So I had to add 33 to keep the value the same. Then 4 + 33 = 37

  34. anonymous
    • one year ago
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    This is what I found in a table for the right integral

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  35. anonymous
    • one year ago
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    a = -1, b = 6, c = 3, and n = 2

  36. anonymous
    • one year ago
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    So once you fill in all the constants and do what looks to be an obnoxious amount of multiplication, all that's left is to solve the integral on the right

  37. anonymous
    • one year ago
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    |dw:1434691058829:dw|

  38. anonymous
    • one year ago
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    The top one

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  39. dessyj1
    • one year ago
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    is what you just sent a suggestion? can i just use the previous formula?

  40. anonymous
    • one year ago
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    you need them both. If you notice, the first one has and integral toward the end of it. The second one is just the solution to that integral

  41. dessyj1
    • one year ago
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    would the integral at the end of the first be multiplied to its preceding equations?

  42. anonymous
    • one year ago
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    Just the part between the plus and integral sign. I personally wouldn't attempt to simplify it. Put the solution to the integral in brackets and write that part outside

  43. dessyj1
    • one year ago
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    so the jumbled mess behind the integral is kind of like a constant?

  44. anonymous
    • one year ago
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    which part? or which integral?

  45. dessyj1
    • one year ago
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    you showed me the first formula

  46. dessyj1
    • one year ago
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    and behind the integrand of the integral at the end is a constant.

  47. anonymous
    • one year ago
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    oh ok. it's not a constant because the x's are still there. The coefficient of the integral works out to being a constant

  48. anonymous
    • one year ago
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    have you used that second formula yet?

  49. dessyj1
    • one year ago
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    yes i have.

  50. dessyj1
    • one year ago
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    i ended up with \[\sqrt{48}\] as my discriminant, which is not pretty.

  51. anonymous
    • one year ago
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    Use the middle formula. I did b²-4ac when their descriptions are for 4ac-b². Looks like you did the same thing I did.

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  52. dessyj1
    • one year ago
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    it shows b^2-4ac though...

  53. anonymous
    • one year ago
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    The discriminant is still the same for the purpose of the integral. |dw:1434692939744:dw|

  54. anonymous
    • one year ago
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    Above I told you to use the formula with arctan, when we should use the one with ln. Just wanted to make sure we're using the same one

  55. dessyj1
    • one year ago
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    oh okay.

  56. dessyj1
    • one year ago
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    i used the onw with the ln

  57. dessyj1
    • one year ago
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    yeah thats what i got as well.

  58. anonymous
    • one year ago
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    ok cool

  59. dessyj1
    • one year ago
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    + the squarert

  60. dessyj1
    • one year ago
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    |dw:1434693216773:dw|

  61. anonymous
    • one year ago
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    yes

  62. dessyj1
    • one year ago
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    that is not the answer for the right integral is it?

  63. dessyj1
    • one year ago
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    it still have to be multiplies by the sum in the first formula right? and by 37

  64. anonymous
    • one year ago
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    no the answer is the whole thing. |dw:1434693381363:dw|

  65. dessyj1
    • one year ago
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    Okay.

  66. dessyj1
    • one year ago
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    |dw:1434693703858:dw|

  67. dessyj1
    • one year ago
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    |dw:1434693767672:dw|

  68. dessyj1
    • one year ago
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    I have a calc 1 final in about 13 hours and i need to sleep.

  69. dessyj1
    • one year ago
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    Thank you for all your help. I really appreciate it.

  70. anonymous
    • one year ago
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    you're welcome. good luck! You should be fine. Any prof who puts something this insane on a final is a sadist.

  71. anonymous
    • one year ago
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    \[\frac{ 37(-2x+6) }{ -48(-x^2+6x+3) }+\frac{ 37 }{ 24\sqrt{48} }\ln|\frac{ -2x+6+\sqrt{48} }{ -2x+6-\sqrt{48} }|+\frac{ 11 }{ 2\sqrt{-x^2+6x+3}}+C\]

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