ksaimouli
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ksaimouli
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\[\int\limits_{}\frac{ 2x^3 }{ x^2-4 }\]
ksaimouli
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i should do the long division right
ksaimouli
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i got \[x^2-4+\frac{ 8x }{ 2x }\]
ksaimouli
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@zepdrix
SithsAndGiggles
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You're right about doing long division first, but it doesn't look like you did it right.
ksaimouli
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|dw:1361917090294:dw|
SithsAndGiggles
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|dw:1361917028271:dw|
SithsAndGiggles
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Right, so by long division you get
\[2x+\frac{8x}{x^2-4}\]
ksaimouli
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so should i make the whole function as|dw:1361917417172:dw|
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.
ksaimouli
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right here i could use u substitution right
SithsAndGiggles
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Or that, yes. A sub might actually shorten the work, so I'd suggest that.
ksaimouli
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i got \[x^2+4\ln(x^2-4)\]
ksaimouli
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+C
ksaimouli
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is this right @SithsAndGiggles
SithsAndGiggles
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Yes that's right, though I would replace the parentheses with absolute value bars.
ksaimouli
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ya i dont know how to use that in equation so put the ()
ksaimouli
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thx
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 (|)
ksaimouli
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|2x+1|
ksaimouli
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ya i got that thx