Derivate ∫ln(x^1/2) dx, from (x^2)/2 to x^2. Can you show the steps and logic behind

Let F(x) = ∫ln(t^(1/2)) dt, from 1 to x

By the fundamental theorem of calculus, we have:

F'(x) = ln(x^(1/2))

The function presented, given the bounds of x^2/2 to x^2, is going to be:

G(x) = F(x^2) - F(x^2/2)

To find the derivative of G, we use chain rule:

G'(x) = F'(x^2) * 2x - F'(x^2/2) * x
= ln((x^2)^(1/2)) * 2x - ln((x^2/2)^(1/2)) * x
= ln|x| * 2x - ln(|x| / 2^(1/2)) * x ............... <---- This is the closest to the answer in the book
= ln|x| * 2x - (ln|x| - ln(2^(1/2))) * x
= ln|x| * 2x - (ln|x| - (1/2)ln(2)) * x
= ln|x| * 2x - ln|x| * x + (1/2)ln(2)) * x
= ln|x| * x + (1/2)ln(2)) * x

Hope that helps!

EDIT: The book drops the absolute value signs, which is correct if you consider x > 0 for you domain (which makes a lot of sense now that I think about it, because x^(1/2) has a domain of x >= 0).

I believe the more fundamental question is WHY does the integration by parts work instead of the MECHANISM.

I see the logic now.

Int (u*v') = u*v - Int(v*u') is telling you in disguise that (uv)' = u*v' + v*u' and integrating it gives you
uv = Int(u*v' + v*u') = Int(u*v') + Int(v*u') and then you get the desired rule. Interesting. Googling is what helps.

Ln (x^ 1/2) = 1/2 Ln x

Int [Ln u du ] = u (Ln u - 1) +c %%% Integration by parts

Answer:= 1/2 [ x^2 (Ln x^2 -1) - (x^2 /2) (Ln (x^2 /2) -1)]

and simplify it!!!

for integration by part look at
http://www.physicsforums.com/showthread.php?t=124028