# Eliminating ln in \INT questions

Take this for example:

(x - 2) - 4/(x-2) ln |x - 2| + C

( Which I derived from:  {  \INT x(x-4)/(x-2)² . dx,  u = x - 2  } )

How could I get rid of that ln ?
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Asked:
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Commented:
Hi again,
I meant that the result of the integration did seem incorrect at first glance as it was with the other INTs in that thread.
Let's try to evaluate it myself

INT x(x-4)/(x-2)² . dx,  u = x - 2

x =  u+2
dx=du

So we have:
INT ((u+2)(u-2)/u^2)du = INT ((u^2-4)/u^2)du = INT (1-4*u^(-2))du = u - 4*(-1)*u^(-1) + C = u + 4/u + C = (x - 2) + 4/(x - 2) + C

Is this the correct answer? I've not received ln here.
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Author Commented:
Hm. Close:

x + 4/(x - 2) + C

(Perhaps the book is wrong? ...)

Ta.
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Commented:
nope, it's the same.
the difference is only -2 but this is not a problem since it is a constant: C and C - 2 could be considered equivalent since during the integration they will both disapper.
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Author Commented:
Oh yeah. :)

Thanks
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