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cherrilyn

  • 5 years ago

evaluate the integral

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

  2. anonymous
    • 5 years ago
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    You have to give us a hint. Which part of the book are you at?

  3. cherrilyn
    • 5 years ago
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    hehe. partial functions

  4. anonymous
    • 5 years ago
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    \[LHS =A/(X-1) +B(x+1)\]

  5. anonymous
    • 5 years ago
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    B/(x+1)

  6. cherrilyn
    • 5 years ago
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    umm.... wouldn't it be Ax+B/x^2+x? or are you skipping steps

  7. anonymous
    • 5 years ago
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    Oh, I haven't done these in a while, may be you should start it.

  8. anonymous
    • 5 years ago
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    Oh, the bottom is not the difference of two squares, I misread it.

  9. cherrilyn
    • 5 years ago
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    can you only use long division if the denominator is the difference of two squares?

  10. anonymous
    • 5 years ago
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    \[LHS =A/x + B/(x-1)\]

  11. cherrilyn
    • 5 years ago
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    are you sure?

  12. anonymous
    • 5 years ago
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    \[x ^{2}-x=x(x-1)\]

  13. cherrilyn
    • 5 years ago
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    okay so what should I do next

  14. anonymous
    • 5 years ago
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    You can handle it now. It is cut up in small bites that you can chew on.

  15. cherrilyn
    • 5 years ago
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    I got A = -1 and B =1 but when I find the integral my denominator = 0 :/

  16. anonymous
    • 5 years ago
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    what , just equals -ln(x) +ln(x-1) +C

  17. anonymous
    • 5 years ago
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    can be simplifed as ln [ (x-1)/(x) ] +C

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