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# Eliminating ln in \INT questions

Posted on 2006-04-18
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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 ?
0
Question by:InteractiveMind

LVL 11

Accepted Solution

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

0

LVL 25

Author Comment

Hm. Close:

x + 4/(x - 2) + C

(Perhaps the book is wrong? ...)

Ta.
0

LVL 11

Expert Comment

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.
0

LVL 25

Author Comment

Oh yeah. :)

Thanks
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