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cherrilyn

  • 5 years ago

Evaluate the integral ; integral of (x^2+x+3)dx/ (x-1)^3

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  1. cherrilyn
    • 5 years ago
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    \[\int\limits_{}^{} (x ^{2} + x + 3) dx / (x-1)^{3}\]

  2. cherrilyn
    • 5 years ago
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    A/x-1 + B/(x-1)^2 + C/(x-1)^3... then

  3. anonymous
    • 5 years ago
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    http://www.youtube.com/watch?v=qm3Gg42CKXI&feature=related

  4. anonymous
    • 5 years ago
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    You're doing good, you want to know how to finish?

  5. anonymous
    • 5 years ago
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    (x2+x+3)dx/(x−1)3 = A/x-1 + B/(x-1)^2 + C/(x-1)^3

  6. anonymous
    • 5 years ago
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    Multiply through by \[(x-1)^{3}\]

  7. cherrilyn
    • 5 years ago
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    don't I have to get rid of the fractions first ?

  8. anonymous
    • 5 years ago
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    Multiplying by \[(x-3)^{3}\] is a way of getting rid of the fractions.

  9. cherrilyn
    • 5 years ago
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    jk, thats what it does haha YEAH

  10. cherrilyn
    • 5 years ago
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    now find out the variables right?

  11. cherrilyn
    • 5 years ago
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    if x = 1, c = 5 but how do I find out the other variables

  12. anonymous
    • 5 years ago
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    You substitute for 5 for C into the line \[x ^{2}+x+3=A(x-1)^{2}+B(x-1)+C\]

  13. anonymous
    • 5 years ago
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    and expand

  14. cherrilyn
    • 5 years ago
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    how do you expand ?

  15. anonymous
    • 5 years ago
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    Multiply threw with B, expand \[(x-1)^{2}\]

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