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Need help integrating x^3/(x-2)^2.
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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.
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)
I'm not sure if I'm doing this right so far. :(
Please help.
thats right, but its supposed to be 8/(x-2)^2 not 16
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and one more thing, 12/(x-2)
I'm not sure how you got those. Are they a result of long division?
they are the result of long division, lalaly is right.
so you have (int) x + (int) 12/(x-2) + (int) 8/(x+2)^2 + (int) 4
\[x^2/2+12\ln(x-2)-8/(x-2)+4x+C\]
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when integrated you get the above
Ok. Thanks!
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