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anonymous

  • 4 years ago

Need help integrating x^3/(x-2)^2.

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  1. anonymous
    • 4 years ago
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    I started with long division and got x + 4 as the quotient and 12x - 16 as th remainder. I'm not sure if thats correct.

  2. anonymous
    • 4 years ago
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    Then I wrote it as the sum of the integral of each term. So like, (int) x + (int) 4 + (int)(12/(x-2)^2) - (int) (16/(x-2 )^2)

  3. anonymous
    • 4 years ago
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    I'm not sure if I'm doing this right so far. :(

  4. anonymous
    • 4 years ago
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    Please help.

  5. lalaly
    • 4 years ago
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    thats right, but its supposed to be 8/(x-2)^2 not 16

  6. lalaly
    • 4 years ago
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    and one more thing, 12/(x-2)

  7. anonymous
    • 4 years ago
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    I'm not sure how you got those. Are they a result of long division?

  8. anonymous
    • 4 years ago
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    they are the result of long division, lalaly is right.

  9. anonymous
    • 4 years ago
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    so you have (int) x + (int) 12/(x-2) + (int) 8/(x+2)^2 + (int) 4

  10. anonymous
    • 4 years ago
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    \[x^2/2+12\ln(x-2)-8/(x-2)+4x+C\]

  11. anonymous
    • 4 years ago
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    when integrated you get the above

  12. anonymous
    • 4 years ago
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    Ok. Thanks!

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