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baldymcgee6

  • 3 years ago

Integral by partial fraction?

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  1. baldymcgee6
    • 3 years ago
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    |dw:1358736744745:dw|

  2. abb0t
    • 3 years ago
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    \[4x = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }\]

  3. baldymcgee6
    • 3 years ago
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    thank you that helps alot.. how do you come up with that?

  4. baldymcgee6
    • 3 years ago
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    and why is it 4x = ......

  5. baldymcgee6
    • 3 years ago
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    if it is quadratic in denominator in is linear in numerator?

  6. abb0t
    • 3 years ago
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    4x is the numerator. and when it's quadratic it follows the form Ax+B

  7. baldymcgee6
    • 3 years ago
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    so now we solve all the a,b,c,d,e?

  8. baldymcgee6
    • 3 years ago
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    how do we solve for all those?

  9. abb0t
    • 3 years ago
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    You get rid of the denominators.

  10. abb0t
    • 3 years ago
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    \[4x = A(x^2+1)^2(x+1) + Bx+C(x+1)(x^2+1) + Dx+E(x+1)\]

  11. zepdrix
    • 3 years ago
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    \[\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }\] The left should look like this before you get rid of the denominators :) The rest of that setup looks great though.

  12. baldymcgee6
    • 3 years ago
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    ahh, thank you zepdrix, was a little confused

  13. baldymcgee6
    • 3 years ago
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    I think i can handle it from here, I just have hard time setting the partial fractions up. any tips?

  14. abb0t
    • 3 years ago
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    Give me a minute to write it all out for you, how in general it works

  15. baldymcgee6
    • 3 years ago
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    take your time, i'll work on this problem, thanks so much

  16. baldymcgee6
    • 3 years ago
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    \[4x = A{\left( {{x^2} + 1} \right)^2} + Bx + C\left( {{x^2} + 1} \right)\left( {x + 1} \right) + Dx + E\left( {x + 1} \right)\]

  17. zepdrix
    • 3 years ago
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    When you multiply through by that denominator you should get, \[\large \begin{align*} 4x = &A(x^2+1)^2\\&+(Bx+C)(x^2+1)(x+1)\\&+(Dx+E)(x+1) \end{align*}\] There may have been a small boo boo when abbot multiplied that out, so I just wanted to check.

  18. zepdrix
    • 3 years ago
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    No I guess it was fine, just don't forget the brackets on top.

  19. zepdrix
    • 3 years ago
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    You can easily solve for A by plugging in x=-1. That should give you a 0 on the last 2 large terms. For the rest of them... you have to deal with a nasty squared X term. So you'll have to expand aaaaaaall of that out :( kinda sucks...

  20. baldymcgee6
    • 3 years ago
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    @abb0t did have an extra (x+1) on the first partial..

  21. zepdrix
    • 3 years ago
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    Oh i see c:

  22. abb0t
    • 3 years ago
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    FOR: \[ax+b \] the term in partial fraction decomposition is: \[\frac{ A }{ ax+b }\] FOR: \[(ax+b)^n\] the term in partial fraction decomposition is: \[\frac{ A_1 }{ ax+b }+\frac{ A_2 }{ (ax+b)^2 }+...+\frac{ A_n }{ (ax+b)^n }\] where n = 1, 2, 3, ... FOR: \[ax^2+bx+c\] the term in partial fraction decomposition is: \[\frac{ Ax+B }{ ax^2+bx+c }\] and FINALLY, for: \[(ax^2+bx+c)^n\] the partial fration decomposition is: \[\frac{ A_1x+B_1 }{ ax^2+bx+c }+\frac{ A_2x+B_2 }{ (ax^2+bx+c)^2 }+...+\frac{ A_nx+B_n }{ (ax^2+bx+c)^n }\] where same as previously, n = 1, 2, 3...

  23. zepdrix
    • 3 years ago
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    Oh boy... abbot gettin all fancy -_- lol

  24. abb0t
    • 3 years ago
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    haters gon hate :P jk :)

  25. zepdrix
    • 3 years ago
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    XD

  26. baldymcgee6
    • 3 years ago
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    haha that helps so much... my prof didn't provide us with this info.

  27. abb0t
    • 3 years ago
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    I think it's because you should have learned this in I think algebra or pre-cal. That's why they usually don't go over partial fraction decomp :P

  28. baldymcgee6
    • 3 years ago
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    :( we had a whole lesson on partial fractions and he seemed to miss "how to split it up"

  29. zepdrix
    • 3 years ago
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    Need help finding any of the unknown constants? Or you tearing through this beast just fine? :O

  30. baldymcgee6
    • 3 years ago
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    haha i had to reboot my comp, but i will slay him now :)

  31. baldymcgee6
    • 3 years ago
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    oh man, how do i solve for the constants?? I got A = -1, but i'm not sure what to use for x to solve for the others

  32. baldymcgee6
    • 3 years ago
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    or should I expand everything first like you said?

  33. zepdrix
    • 3 years ago
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    Yah you have to expand them all out :c sucks.

  34. zepdrix
    • 3 years ago
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    I think.

  35. baldymcgee6
    • 3 years ago
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    haha, well, i will give it a shot.

  36. baldymcgee6
    • 3 years ago
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    too bad wolfram wont do it for me

  37. baldymcgee6
    • 3 years ago
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    I got something ridiculously long

  38. zepdrix
    • 3 years ago
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    I got A=-2 I think... Hmm

  39. zepdrix
    • 3 years ago
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    Oh you square the square thingy, nevermind -1 sounds correct :)

  40. baldymcgee6
    • 3 years ago
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    wait i think it's positive 1 now..

  41. baldymcgee6
    • 3 years ago
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    oh nope.. -1

  42. baldymcgee6
    • 3 years ago
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    haha, second guessing,

  43. baldymcgee6
    • 3 years ago
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    I have no idea how to solve the other constants.

  44. zepdrix
    • 3 years ago
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    \[\large \begin{align*}4x=&-(x^4+2x^2+1)\\&+(Bx^4+Bx^3+Bx^2+Bx\\&+Cx^3+Cx^2+Cx+C)\\&+(Dx^2+Dx+Ex+E)\end{align*}\]

  45. zepdrix
    • 3 years ago
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    Did you get something like this? I ran through it a little quickly, so I hope I didn't make any mistakes.

  46. baldymcgee6
    • 3 years ago
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    yep thats what i got.

  47. zepdrix
    • 3 years ago
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    From here, we'll EQUATE LIKE TERMS. So see how we have \(0x^4\) on the left? we'll set that equal to all the \(x^4\)'s on the right.\[\large 0=-x^4+Bx^4\]Dividing both sides by x gives us,\[\large 0=-1+B\]

  48. zepdrix
    • 3 years ago
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    I should have written \(0x^4\) on the left to stay consistent. Oh well..

  49. baldymcgee6
    • 3 years ago
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    I don't understand what you mean by 0x^4 and where it should be.

  50. baldymcgee6
    • 3 years ago
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    4x is on the left...

  51. zepdrix
    • 3 years ago
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    You equate all terms that of the same degree. There is a 4x on the left. But that is X to the FIRST degree. We'll equate that with all the of FIRST degree X's on the right.

  52. baldymcgee6
    • 3 years ago
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    ohhhhhh

  53. zepdrix
    • 3 years ago
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    You have NO 4th degree x's on the left. So you equate those with the 4th degree x's on the right.

  54. baldymcgee6
    • 3 years ago
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    :) i get ya now.

  55. zepdrix
    • 3 years ago
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    To be honest the hardest part of the problem is over now :) Now we just need to be careful, matching up the orders correctly so we get the right constants.

  56. baldymcgee6
    • 3 years ago
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    ha ok. I've never heard of equating the like terms before..

  57. zepdrix
    • 3 years ago
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    Ya none of my teachers have really explained that step to me very well... So I wouldn't know how to justify it :P

  58. baldymcgee6
    • 3 years ago
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    |dw:1358740342133:dw|

  59. baldymcgee6
    • 3 years ago
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    |dw:1358740410506:dw|

  60. zepdrix
    • 3 years ago
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    That's kinda funny... I don't think we'll even end up using the 4x to find our unknown constants lol

  61. baldymcgee6
    • 3 years ago
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    we dont need it to solve for E?

  62. zepdrix
    • 3 years ago
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    Oh I suppose we could! My bad. I was thinking of matching up the 0th degree X's to find E. But we can match up the 1st degree X's also.

  63. zepdrix
    • 3 years ago
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    Since there's an Ex and also an E term.

  64. zepdrix
    • 3 years ago
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    So either will work.

  65. baldymcgee6
    • 3 years ago
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    oh, 0th degree would be easier... then |dw:1358740648463:dw|

  66. zepdrix
    • 3 years ago
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    looks good c: I came up with the same values.

  67. baldymcgee6
    • 3 years ago
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    to clarify: A = -1 B = 1 C = -1 D = 2 E = 2

  68. zepdrix
    • 3 years ago
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    mhm

  69. baldymcgee6
    • 3 years ago
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    \[\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ -1 }{ (x+1) } + \frac{ x-1 }{ (x^2+1) } + \frac{ 2x+2 }{ (x^2+1)^2 }\]

  70. zepdrix
    • 3 years ago
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    So that's the contents of our new integral? k looks good.

  71. baldymcgee6
    • 3 years ago
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    and now i can integrate it right?

  72. zepdrix
    • 3 years ago
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    Yes, we might have to do a little work to integrate some of the pieces. Like, we'll have to write the last 2 terms... umm.. we'll need to split up the fractions.

  73. zepdrix
    • 3 years ago
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    We're going to end up with a bunch of natural logs and arctangents i think.

  74. abb0t
    • 3 years ago
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    So, what did I miss?

  75. baldymcgee6
    • 3 years ago
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    |dw:1358740971154:dw|

  76. zepdrix
    • 3 years ago
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    \[\int\limits -\frac{1}{x+1} + \color{green}{\frac{x}{x^2+1}-\frac{1}{x^2+1}} + \color{cornflowerblue}{\frac{2x}{(x^2+1)^2}+\frac{2}{(x^2+1)^2}}dx\]

  77. zepdrix
    • 3 years ago
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    Understand what I mean about splitting up the fractions?

  78. baldymcgee6
    • 3 years ago
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    yep, works for me!

  79. zepdrix
    • 3 years ago
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    Yay! So integrate! c: I'm not actually sure what the last term is going to give us. I can't think of it off the top of my head lol

  80. abb0t
    • 3 years ago
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    I think it's arctan?

  81. zepdrix
    • 3 years ago
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    naw, see how the bottom is being squared? It requires a trig sub and then some fussing with D':

  82. baldymcgee6
    • 3 years ago
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    ahh, i'll figure it out from here guys, thanks for the help!!!!

  83. baldymcgee6
    • 3 years ago
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    wish i could give you both medals

  84. zepdrix
    • 3 years ago
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    yay team \c:/

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