Question

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

Answer 1

$\int \frac{x-1}{\sqrt{x^2-1}}dx=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx$ .........(1)
For $\int \frac{x}{\sqrt{x^2-1}}dx$, let $x^2-1=t\Rightarrow 2xdx=dt$
$\therefore \int \frac{x}{\sqrt{x^2-1}}dx=\frac{1}{2}\int \frac{dt}{\sqrt{t}}$
$=\frac{1}{2}\int t^{\frac{1}{2}}dt=\frac{1}{2}\left[2t^{\frac{1}{2}}\right]=\sqrt{t}=\sqrt{x^2-1}$
From (1), we obtain
$\int \frac{x-1}{\sqrt{x^2-1}}dx=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx,[\because \int \frac{1}{\sqrt{x^2-a^2}}dt=\log |x+\sqrt{x^2-a^2}|]$
$=\sqrt{x^2-1}-\log |x+\sqrt{x^2-1}|+C$