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Integration that leads to a inverse trig identity

  1. Dec 7, 2008 #1
    Ok, so a teacher showed an example in class awhile back. So im going over my notes right now, and I don't understand a certain part of the problem.

    Also im new to the forums and its my first time posting here, so please push me in the right direction if i make a mistake.

    integration of (x-2)(1/2)/(x+2)

    the solution is: 2(x-2).5 - 4tan-1((x-2).5/2) + C

    Basically what i tried doing, is:

    u2 = x-2
    u = (x-2).5
    2u du = dx

    which after a few steps leads me to:

    2u - 8(integration of (u2 +4)-1)

    After this i stop understanding the problem a little...
    From here i sub the root x-2 back in as u and u2 as x-2, which gives me:

    2(x-2).5 - 8ln|x+2| which is wrong i think....

    The way the example continues is as such:
    u2 = 4z2
    u = 2z
    du = 2 dz

    2u - ((8)(2)/4) integration of dz/ (z2 +1)

    which gives the answer given above...

    Basically i dont understand why we want to sub in another letter for u2, and why i cant get the same answer when i sub in the (x-2).5 earlier.
  2. jcsd
  3. Dec 7, 2008 #2


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    Welcome to PF!

    Hi Minutemade! Welcome to PF! :smile:
    ah … you forgot to change the du as well. :wink:
    You don't really need to …

    most people wouldn't …

    but it does stop you making a mistake if you can't remember the general formula! :smile:
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