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

We need to evaluate the integral: $\int \frac{x-1}{\sqrt{x^{2}-1}} dx$
Split the integral into two parts:
$\int \frac{x-1}{\sqrt{x^{2}-1}} dx = \int \frac{x}{\sqrt{x^{2}-1}} dx - \int \frac{1}{\sqrt{x^{2}-1}} dx$ ............ $(1)$
For the first part,let $I_1 = \int \frac{x}{\sqrt{x^{2}-1}} dx$.
Let $x^{2}-1 = t$,then $2x dx = dt$,or $x dx = \frac{1}{2} dt$.
$I_1 = \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot 2t^{1/2} = \sqrt{t} = \sqrt{x^{2}-1}$.
For the second part,we use the standard integral formula:
$\int \frac{1}{\sqrt{x^{2}-a^{2}}} dx = \log |x + \sqrt{x^{2}-a^{2}}| + C$.
Here $a = 1$,so $\int \frac{1}{\sqrt{x^{2}-1}} dx = \log |x + \sqrt{x^{2}-1}|$.
Combining these results in $(1)$:
$\int \frac{x-1}{\sqrt{x^{2}-1}} dx = \sqrt{x^{2}-1} - \log |x + \sqrt{x^{2}-1}| + C$,where $C$ is an arbitrary constant.