Integrate the function: $\frac{x+2}{\sqrt{x^{2}-1}}$

Let $x+2=A \frac{d}{d x}(x^{2}-1)+B$ ...........$(1)$
$\Rightarrow x+2=A(2 x)+B$
Equating the coefficients of $x$ and constant terms 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,$\int \frac{x+2}{\sqrt{x^{2}-1}} d x=\int \frac{\frac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$
$=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x+\int \frac{2}{\sqrt{x^{2}-1}} d x$ ..........$(2)$
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,$\int \frac{2}{\sqrt{x^{2}-1}} d x = 2 \log |x + \sqrt{x^{2}-1}|$
From equation $(2),$ we obtain:
$\int \frac{x+2}{\sqrt{x^{2}-1}} d x = \sqrt{x^{2}-1} + 2 \log |x + \sqrt{x^{2}-1}| + C$
where $C$ is an arbitrary constant.