$\cfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}$

: Let $I = \int {\cfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}} dx$

$= \int {\cfrac{x}{{\sqrt {{x^2} - 1} }}dx} - \int {\cfrac{1}{{\sqrt {{x^2} - 1} }}dx} = {I_1} - {I_2}$

(say)
Now, ${I_1} = \int {\cfrac{x}{{\sqrt {{x^2} - 1} }}dx}$

Put ${x^2} - 1 = t$ $\Rightarrow$ $2x\,dx\, = dt$

*********************************

$\therefore$ $I = \cfrac{1}{2}\int {\cfrac{{dt}}{{\sqrt t }} = \cfrac{1}{2}} \int {{t^{ - 1/2}}} dt = \cfrac{1}{2} \times \cfrac{{{t^{1/2}}}}{{1/2}} + {C_1}$

$= \sqrt t + {C_1} = \sqrt {{x^2} - 1} + {C_1}$

And ${I_2} = \int {\cfrac{1}{{\sqrt {{x^2} - 1} }}dx} = \log \left| {x + \sqrt {{x^2} - 1} } \right| + {C_2}$

$\therefore$ $I = \sqrt {{x^2} - 1} - \log \left| {x + \sqrt {{x^2} - 1} } \right| + C$

where,$C = {C_1} - {C_2}$