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ksaimouli

by partial frac

  • one year ago
  • one year ago

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  1. ksaimouli
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    \[\int\limits_{}\frac{ 2x^3 }{ x^2-4 }\]

    • one year ago
  2. ksaimouli
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    i should do the long division right

    • one year ago
  3. ksaimouli
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    i got \[x^2-4+\frac{ 8x }{ 2x }\]

    • one year ago
  4. ksaimouli
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    @zepdrix

    • one year ago
  5. SithsAndGiggles
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    You're right about doing long division first, but it doesn't look like you did it right.

    • one year ago
  6. ksaimouli
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    |dw:1361917090294:dw|

    • one year ago
  7. SithsAndGiggles
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    |dw:1361917028271:dw|

    • one year ago
  8. SithsAndGiggles
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    Right, so by long division you get \[2x+\frac{8x}{x^2-4}\]

    • one year ago
  9. ksaimouli
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    so should i make the whole function as|dw:1361917417172:dw|

    • one year ago
  10. SithsAndGiggles
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    You can, but that doesn't mean you should. Your integral has changed from \[\int\frac{2x^3}{x^2-4}dx\] to \[\int\left(2x+\frac{8x}{x^2-4}\right)\;dx\] Integrate term by term, with partial fraction decomp on the second term.

    • one year ago
  11. ksaimouli
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    right here i could use u substitution right

    • one year ago
  12. SithsAndGiggles
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    Or that, yes. A sub might actually shorten the work, so I'd suggest that.

    • one year ago
  13. ksaimouli
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    i got \[x^2+4\ln(x^2-4)\]

    • one year ago
  14. ksaimouli
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    +C

    • one year ago
  15. ksaimouli
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    is this right @SithsAndGiggles

    • one year ago
  16. SithsAndGiggles
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    Yes that's right, though I would replace the parentheses with absolute value bars.

    • one year ago
  17. ksaimouli
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    ya i dont know how to use that in equation so put the ()

    • one year ago
  18. ksaimouli
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    thx

    • one year ago
  19. SithsAndGiggles
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    You're welcome. By the way, absolute value bars are above the Enter/Return key. Hold shift and click the backslash (\) button to get (|)

    • one year ago
  20. ksaimouli
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    |2x+1|

    • one year ago
  21. ksaimouli
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    ya i got that thx

    • one year ago
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