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Eliminating ln in \INT questions

Posted on 2006-04-18
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Last Modified: 2006-11-18
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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Question by:InteractiveMind
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WelkinMaze earned 1000 total points
ID: 16477028
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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by:InteractiveMind
ID: 16477245
Hm. Close:

   x + 4/(x - 2) + C

(Perhaps the book is wrong? ...)

Ta.
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by:WelkinMaze
ID: 16477270
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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by:InteractiveMind
ID: 16477420
Oh yeah. :)

Thanks
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