Integration help

Check your integrand in u on the penultimate line. You know the result is ln(), so that must be a ....
Looks ok otherwise. I would have simply done dx/du = 2u though. Also remember its a definite integral, so you have limits.

1/2 x^-1

Not sure how you get that?
Just carefully apply your substitution to the original integral.

Oh so it should be 2f (2+u)^-1 du

Yes, so just integrate.
Remember the original limits (definite integral) as well.

So 2ln|x+2|

Sort of, its "u" not "x" . Then apply the definite integral limits.
Note the two multiplier can brought inside the log as a power using the usual rules.
This gives something similar (not identical) to the expressions you give at the end of #2.

ln|5+2 root 6 /2|

I’ve subbed the limits into u= x^1/2
Then ln(2+root 6)^2 - ln(4)
ln (( 2+root 6)^2)/4

Thats different from the previous expression and still not correct.
I still don't know how you get those terms.

Oh I think it because I subbed back the x terms instead of leaving it in U like I was supposed to

Try and do it carefully and upload your full working if necessary?

I got ln16

Last try, can you upload your full working.

Being negative, you could guestimate the integral is ~0.3 as the width is 2 (4-2) and the height is ~1/6. So ln(16) is way out. However, it is part of the answer.

So the limits are different from the OP? They were 2 to 4, now they're 0 to 36? ln(16) is correct for these limits.