ksaimouli
  • ksaimouli
by partial frac
Mathematics
jamiebookeater
  • jamiebookeater
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ksaimouli
  • ksaimouli
\[\int\limits_{}\frac{ 2x^3 }{ x^2-4 }\]
ksaimouli
  • ksaimouli
i should do the long division right
ksaimouli
  • ksaimouli
i got \[x^2-4+\frac{ 8x }{ 2x }\]

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

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