anonymous
  • anonymous
Need help integrating x^3/(x-2)^2.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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.
anonymous
  • anonymous
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)
anonymous
  • anonymous
I'm not sure if I'm doing this right so far. :(

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anonymous
  • anonymous
Please help.
lalaly
  • lalaly
thats right, but its supposed to be 8/(x-2)^2 not 16
lalaly
  • lalaly
and one more thing, 12/(x-2)
anonymous
  • anonymous
I'm not sure how you got those. Are they a result of long division?
anonymous
  • anonymous
they are the result of long division, lalaly is right.
anonymous
  • anonymous
so you have (int) x + (int) 12/(x-2) + (int) 8/(x+2)^2 + (int) 4
anonymous
  • anonymous
\[x^2/2+12\ln(x-2)-8/(x-2)+4x+C\]
anonymous
  • anonymous
when integrated you get the above
anonymous
  • anonymous
Ok. Thanks!

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