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ksaimouli

  • 2 years ago

by partial frac

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

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

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

  4. ksaimouli
    • 2 years ago
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    @zepdrix

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

  6. ksaimouli
    • 2 years ago
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    |dw:1361917090294:dw|

  7. SithsAndGiggles
    • 2 years ago
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    |dw:1361917028271:dw|

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

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

  10. SithsAndGiggles
    • 2 years ago
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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.

  11. ksaimouli
    • 2 years ago
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    right here i could use u substitution right

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

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

  14. ksaimouli
    • 2 years ago
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    +C

  15. ksaimouli
    • 2 years ago
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    is this right @SithsAndGiggles

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

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

  18. ksaimouli
    • 2 years ago
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    thx

  19. SithsAndGiggles
    • 2 years ago
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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 (|)

  20. ksaimouli
    • 2 years ago
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    |2x+1|

  21. ksaimouli
    • 2 years ago
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    ya i got that thx

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