Inverse hyperbolic function integral

[iamalexalright]

lets see here, I am trying to integrate this(and sorry, i don't know how to use the symbols - ill use '{' as my integral sign):
6
{ 1 / (t^2 - 9) ^ .5
4

so, considering { 1 / (x^2 - 1) ^ .5 = inverse cosh(x) i did:

6
(1/3) { 1 / ((t / 3) ^ 2 - 1) ^ .5
4

then i took the integral so i got:

6
(1/3) inverse cosh(t / 3) |
4

so then, considering inverse cosh(u) = ln(u + (u - 1) ^ .5)

i made my equation into:

6
(1/3) ln((t/3) + ((t/3) - 1) ^ .5)) |
4

and then computing that i come up with .1738641452 but the books answer is :

ln ((6 + 3(3)^.5)/(4 + 7^.5)) which is not the same as mine(plus it is 'prettier' in the sense that it looks like i could get that answer without a calculator which is how I am supposed to be doing it).

any help or suggestions? sorry about the readability

 

[interested_learner]

LaTex

Go to the Math and Science Tutorials section. Then to the Introducing LaTex Math Typesetting tutorial. It is easy to use. You can't really expect volunteers to help you unless you make it easy for them to read you question.

 

[dextercioby]

\int_{4}^{6}\frac{dx}{\sqrt{x^{2}-9}}=\allowbreak \ln 3+\ln \left( 2+\sqrt{3}\right) -\ln \left( 4+\sqrt{7}\right)

To get there, make the substitution

x=3\cosh t

and pay attention when you evaluate \cosh^{-1} 2 and \cosh^{-1} 4/3.

Daniel.