$\displaystyle \int_1^2 \frac{\sqrt{x^2-1}}{x}dx$

$\displaystyle x = sec\theta$

$\displaystyle \sqrt{x^2-1} = tan\theta$

$\displaystyle dx = sec\theta*tan\theta$

$\displaystyle x = 2 ; \theta = \pi/3$

$\displaystyle x = 1 ; \theta = 2\pi$

$\displaystyle \int_{2\pi}^{\pi/3} \frac{tan\theta}{sec\theta}sec\theta*tan\theta$

This isnt yielding me the correct result, where am I going wrong?

Thanks