Integral of hyperbolic function

[DryRun]

Homework Statement
$$\int^4_3 \frac{1}{\sqrt{3x^2-6x+1}}\,.dx$$

The attempt at a solution
I complete the square for the quadratic:
$$\sqrt{3x^2-6x+1} \\=\sqrt{3(x^2-2x+\frac{1}{3})} \\=\sqrt 3 \times \sqrt{(x-1)^2-\frac{2}{3}}$$
$$\int^4_3 \frac{1}{\sqrt{3x^2-6x+1}}\,.dx \\=\frac{1}{\sqrt 3}\int^4_3 \frac{1}{\sqrt{(x-1)^2-\frac{2}{3}}}\,.dx \\=\frac{1}{\sqrt 3} \left[\cosh^{-1}\frac{\sqrt 3(x-1)}{\sqrt 2}\right]^{x=4}_{x=3} \\=\frac{1}{\sqrt 3} \left[\cosh^{-1}\frac{3\sqrt 3}{\sqrt 2}-\cosh^{-1}\sqrt 6\right]$$
I already simplified it but it doesn't agree with the final answer:
$$\frac{1}{\sqrt 3}\ln \left(\frac{15+\sqrt {219}}{12+\sqrt {138}}\right)$$

 

[tiny-tim]

hi sharks!

cosh^-1x = ln(e^cosh-1x)

= ln(cosh(cosh^-1x) + sinh(cosh^-1x))

= ln (x + √(x² - 1)) ​

 

[DryRun]

Hi tiny-tim

The above expression is what i used to expand:
$$\frac{1}{\sqrt 3} \left [\cosh^{-1}\frac{3\sqrt 3}{\sqrt 2}-\cosh^{-1}\sqrt 6\right] \\=\frac{1}{\sqrt 3} \ln \left[\frac {3\sqrt 3+5}{\sqrt{12}+\sqrt{10}}\right]=0.4625025064$$
But if i evaluate the answer from my notes, i get:
$$\frac{1}{\sqrt 3}\ln \left (\frac{15+\sqrt {219}}{12+\sqrt {138}}\right)=0.1310541888$$
Since the answers are not the same, I'm thinking that maybe the answer in my notes is wrong?

 

[tiny-tim]

well that's obviously …
$$\frac{1}{\sqrt 3} \left[\cosh^{-1}\frac{\sqrt 3(x+1)}{\sqrt 2}\right]^{x=4}_{x=3}$$

… either the question or the answer is wrong