Prove

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

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

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

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

$=\frac{1}{2}\left[2 t^{\frac{1}{2}}\right]$

$=\sqrt{t}$

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

From (1), we obtain

$\int \frac{x-1}{\sqrt{x^{2}-1}} d x=\int \frac{x}{\sqrt{x^{2}-1}} d x-\int \frac{1}{\sqrt{x^{2}-1}} d x$  $\left[\int \frac{1}{\sqrt{x^{2}-a^{2}}} d t=\log \left|x+\sqrt{x^{2}-a^{2}}\right|\right]$

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