Integration that leads to a inverse trig identity

[Minutemade]

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

Also I am 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 don't understand why we want to sub in another letter for u², and why i can't get the same answer when i sub in the (x-2)^.5 earlier.

 

[tiny-tim]

Welcome to PF!

Hi Minutemade! Welcome to PF!

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...

ah … you forgot to change the du as well.

Basically i don't understand why we want to sub in another letter for u²…

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!