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El_Arrow

  • one year ago

need help with integral problem

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  1. anonymous
    • one year ago
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    ?

  2. anonymous
    • one year ago
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    What's the question/problem

  3. El_Arrow
    • one year ago
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    i am typing it man

  4. anonymous
    • one year ago
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    ok

  5. El_Arrow
    • one year ago
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    \[\int\limits_{?}^{?}\frac{ 3x^3-3x^2+6x-6}{ (x^2+2) (x-2)^3}\]

  6. El_Arrow
    • one year ago
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    there is the integral and i am using partial fraction

  7. anonymous
    • one year ago
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    Before you do partial fractions, factor the numerator. Nice stuff happens :)

  8. El_Arrow
    • one year ago
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    this is what i have so far |dw:1433371363783:dw|

  9. anonymous
    • one year ago
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    It will dumb down to something better if you can factor the numerator properly.

  10. El_Arrow
    • one year ago
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    so 3(x^3 - x^2 +2x - 2)

  11. geerky42
    • one year ago
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    Factor more.

  12. geerky42
    • one year ago
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    \[x^3 - x^2 + 2x-2 ~~~=~~~ x^2(x-1) + 2(x-1)\]

  13. El_Arrow
    • one year ago
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    oh

  14. anonymous
    • one year ago
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    Yeah, the numerator is a factoring by grouping example.

  15. El_Arrow
    • one year ago
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    so after that what do i do

  16. anonymous
    • one year ago
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    Well, \(x^{2}(x-1) + 2(x-1)\) can factor also.

  17. El_Arrow
    • one year ago
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    in to what?

  18. anonymous
    • one year ago
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    Well, there is a common (x-1) term. So factor out the (x-1) \(x^{2}(x-1) + 2(x-1) = (x^{2} + 2)(x-1)\) You see how that works?

  19. El_Arrow
    • one year ago
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    so its 3(x^2+2)(x-1) right

  20. anonymous
    • one year ago
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    Yep, which means the \(x^{2} + 2\) would cancel

  21. El_Arrow
    • one year ago
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    which leaves you with |dw:1433371872856:dw|

  22. anonymous
    • one year ago
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    Yep. And we can actually avoid partial fractions with that. Because yeah, partial fractions are annoying. \[\int\limits_{}^{}\frac{3x-3}{(x-2)^{3}}dx = \int\limits_{}^{}\frac{ 3x-3 -3 + 3 }{ (x-2)^{3} }dx =\int\limits_{}^{}\frac{ 3x-6 }{ (x-2)^{3} }dx+\int\limits_{}^{}\frac{ 3 }{ (x-2)^{3} }dx\] See what I did there?

  23. El_Arrow
    • one year ago
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    no i am lost

  24. El_Arrow
    • one year ago
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    how did you get -3+3?

  25. anonymous
    • one year ago
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    I distributed the 3 first, obviously. After, I added and subtracted 3 at the same time, which is equivalent to adding 0, so nothing wrong with that. Doing that allows me to split up the numerator. 3x -3 -3 + 3 = 3x-6 + 3 Once I get 3x-6 + 3, I split those into two separate integrals. The 3x-6 is nice because from there I have: \[\int\limits_{}^{}\frac{ 3x-6 }{ (x-2)^{3} }dx + \int\limits_{}^{}\frac{ 3 }{ (x-2)^{3} }dx = 3\int\limits_{}^{}\frac{ (x-2) }{ (x-2)^{3} }dx + 3\int\limits_{}^{}\frac{ 1 }{ (x-2)^{3} }dx\] The + 3 -3 thing is just a neat trick that lets you split up fractions faster. Its basically doing partial fractions but without al the long work.

  26. El_Arrow
    • one year ago
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    so if i did partial fractions i would get the same result?

  27. anonymous
    • one year ago
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    Like for example, if I had this integral: \[\int\limits_{}^{}\frac{ 1 }{ e^{x} + 1 }dx\] I would do that same trick of adding and subtracting the same thing. I would do this: \[\int\limits_{}^{}\frac{1+e^{x} - e^{x}}{e^{x}+1}dx = \int\limits_{}^{}\frac{ 1+e^{x} }{ 1+e^{x} }dx - \int\limits_{}^{}\frac{ e^{x} }{ e^{x}+1 }dx\] So it's a good trick to know for integrals. And yes, it would give you the exact same thing.

  28. El_Arrow
    • one year ago
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    okay but than how would you integrate that?

  29. anonymous
    • one year ago
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    Well, our integral we have: \[3\int\limits_{}^{}\frac{ 1 }{ (x-2)^{2} }dx + 3\int\limits_{}^{}\frac{ 1 }{ (x-2)^{3} }dx\] I just canceled the x-2 in the first integral. But each of these are just u substitution. Let u = x-2

  30. El_Arrow
    • one year ago
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    quick question if i did use partial fractions i would put the 3 outside the integral right

  31. El_Arrow
    • one year ago
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    like this?|dw:1433372913732:dw|

  32. anonymous
    • one year ago
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    Well, putting the 3 outside is preference. I personally dont do it when Im integrating or doing partial fractions or anything, but Ive noticed that a lot of people like moving it outside. I usually keep the 3 in.

  33. anonymous
    • one year ago
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    So if you wished to do partial fractions, I would just have: \[\frac{ 3x-3 }{ (x-2)^{3} } = \frac{ A }{ x-2 }+ \frac{ B }{ (x-2)^{2} } + \frac{ C }{ (x-2)^{3} }\]

  34. El_Arrow
    • one year ago
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    ok i see

  35. El_Arrow
    • one year ago
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    right?|dw:1433373114890:dw|

  36. anonymous
    • one year ago
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    Yes

  37. El_Arrow
    • one year ago
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    so i put 2 for x right?

  38. El_Arrow
    • one year ago
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    for C i got 3 and i dont know what number to put for A or B

  39. anonymous
    • one year ago
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    Yeah, C = 3. Since C = 3, we can do this: \(3x-3 = A(x-2)^{2} + B(x-2) + 3\) \(3x-6 = A(x-2)^{2}+ B(x-2)\) From here, you have the option of distributing everything out and equating powers or we can pick a couple random values of x and solve a system of equations.

  40. El_Arrow
    • one year ago
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    which option is easier?

  41. El_Arrow
    • one year ago
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    which option is easier?

  42. anonymous
    • one year ago
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    Well, foiling out is pretty easy, so I would just do that. If the foiling was annoying, I might do the system route. But you can foil this out and get: \(3x-6 = A(x^{2} -4x + 4) + Bx - 2B\) \(3x-6 = Ax^{2} - 4Ax + 4A + Bx - 2B\) So by equating powers, the left has no \(x^{2}\) terms and the right side has \(Ax^{2}\). This means A has to be 0.

  43. anonymous
    • one year ago
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    Since A is 0, you then would have: \(3x-6 = B(x-2)\) Pick x to be 0 and you get \(-6 = -2B\) so B = 3, which is what we had through the tricky method, we got 3 and 3.

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