Integration- hyperbolic trig can't find my mistake

[Zoe-b]

Homework Statement

(I haven't got the thing to write equations so this is a little difficult to understand sorry!)
Show that the integral of (4x^2 - 1)^-1/2 from 0.625 to 1.3 is equal to 1/2(ln 5 - ln 2)

Homework Equations

my formula book states:
the integral of (x^2 - a^2)^-1/2 is arcosh (x/a) or ln (x + (x^2 - a^2)^1/2)

The Attempt at a Solution

the original integral equals:
1/2 the integral of (x^2 - 1/4)^-1/2
comparing this to the stated result I get a^2 = 1/4
so substituting in limits:
= 1/2 ln(1.3 + (1.3^2 -1/4)^1/2) - 1/2 ln(0.625 + (0.625^2 - 1/4)^1/2)
=1/2 (ln 2.5 - ln 1)

which is incorrect. I just about follow the worked solution, which uses a very slightly different method, but I want to know why this is incorrect because I can't find a mistake in it.. Thanks!

 

[rocomath]

It is correct

\frac 1 2(\ln{\frac 5 2}-\ln 1) =\frac 1 2(\ln 5 - \ln 2)

 

[Zoe-b]

oh yeah.. oops! I didn't realize.
Thank you!