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

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

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

(say)
$\Rightarrow$ $I = {I_1} + {I_2}$

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

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

$\therefore$ ${I_1} = \cfrac{1}{2}\int {\cfrac{{dt}}{t}} = \cfrac{1}{2}\int {{t^{ - 1/2}}} dt = \cfrac{1}{2} \times \cfrac{{{t^{1/2}}}}{{1/2}} = \sqrt t + {C_1} = \sqrt {{x^2} - 1} + {C_1}$

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

….(ii)
From (i), (ii) and (iii),

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

Where, $C = {C_1} + {C_2}$