Eliminating ln in \INT questions

Posted on 2006-04-18
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 ?
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

    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.
    LVL 25

    Author Comment

    Hm. Close:

       x + 4/(x - 2) + C

    (Perhaps the book is wrong? ...)

    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.
    LVL 25

    Author Comment

    Oh yeah. :)


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