Let $I=\int \frac{x+2}{\sqrt{x^2-1}}dx$$\Rightarrow I = I_1 + I_2$ ... (i) Now, $I_1=\int \frac{x}{\sqrt{x^2-1}}dx$ Put $x^2 - 1 = t \Rightarrow 2x dx = dt$ $I_1=\frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} \times \frac{t^{1/2}}{1/2} + C_1$ $= \sqrt{t} + C_1 = \sqrt{x^2-1} + C_1$ ... (ii)and, $I_2=\int \frac{2}{\sqrt{x^2-1}}dx$ $= 2 \log |x + \sqrt{x^2-1}| + C_1$... (iii)From (i), (ii) and (iii), we get $I = \sqrt{x^2-1} + 2 \log |x + \sqrt{x^2-1}| + C$