Prove

Let $x+2=A \frac{d}{d x}\left(x^{2}-1\right)+B$

$\Rightarrow x+2=A(2 x)+B$

Equating the coefficients of x and constant term on both sides, we obtain

$2 A=1 \Rightarrow A=\frac{1}{2}$

$B=2$

From (1), we obtain

$(x+2)=\frac{1}{2}(2 x)+2$

Then, $\int \frac{x+2}{\sqrt{x^{2}-1}} d x=\int \frac{\frac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$

$=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x+\int \frac{2}{\sqrt{x^{2}-1}} d x$    ...(1)

In $\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x$, let $x^{2}-1=t \Rightarrow 2 x d x=d t$

$\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{t}}$

$=\frac{1}{2}[2 \sqrt{t}]$

$=\sqrt{t}$

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

Then, $\int \frac{2}{\sqrt{x^{2}-1}} d x=2 \int \frac{1}{\sqrt{x^{2}-1}} d x=2 \log \left|x+\sqrt{x^{2}-1}\right|$

From equation (2), we obtain

$\int \frac{x+2}{\sqrt{x^{2}-1}} d x=\sqrt{x^{2}-1}+2 \log \left|x+\sqrt{x^{2}-1}\right|+\mathrm{C}$