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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
x+4 + (8x-16)/D
x^2-4x+4 ) x^3
8x - 16
and the Remainder might be able to factor out with the D
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
-2(12)+B = -16
B = 8