## integration- hyperbolic trig.. can't find my mistake

1. The problem statement, all variables and given/known data
(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)

2. Relevant 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)

3. 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!
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 It is correct $$\frac 1 2(\ln{\frac 5 2}-\ln 1) =\frac 1 2(\ln 5 - \ln 2)$$
 oh yeah.. oops! I didn't realise. Thank you!

 Similar discussions for: integration- hyperbolic trig.. can't find my mistake Thread Forum Replies Linear & Abstract Algebra 4 Precalculus Mathematics Homework 2 Calculus & Beyond Homework 4 Calculus & Beyond Homework 2 General Math 7