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I needed to integrate x^3/(x-2)^2. So, I did long division and got x + 4 as the quotient and 12x - 16 as th remainder. So, to get the answer, I was going to take the sum of integrals : ∫x + ∫4 + ∫(12/(x-2)^2) - ∫(16/(x-2)^2). But multiple people have told me that for the last two terms, I should have ∫(12/(x-2)) - ∫(8/(x-2)^2) instead. They said they got it through long division, but I still don't understand how. Please explain. Thanks.
expand the bottom to determine the divisor
and then the results splits apart into the quotient
that has show steps in the division process
x+4 + (8x-16)/D --------------- x^2-4x+4 ) x^3 (x^3-4x^2+4x) ------------- 4x^2-4x (4x^2-16x+16) ------------- 8x - 16 and the Remainder might be able to factor out with the D
8(x-2) --------- (x-2)(x-2)
well, something like that lol; 16-4 = 12 not 8
12x-16 = 2(6x-8) hmm got me, maybe there are 2 ways to express it? might have to work at simplifing further
I don't see why it matters, you get the same result either way. but basically they did partial fractions backwards 12x -16/(x-2)^2 = A/(x-2) + B/(x-2)^2 A(x-2) + B = 12x -16 Ax -2A +B = 12x -16 A=12 -2(12)+B = -16 B = 8
oh, they got the remainder and decomped it?
I think I can handle it from there. :)