\(\int \frac{x - 1}{\sqrt{x^2 + 1}}dx = \int\frac{x}{\sqrt{x^2 + 1}}dx - \int\frac{1}{\sqrt{x^2 + 1}}dx\)
Let \(x^2 + 1 = t^2\)
Then
\(2x \,dx = 2t \,dt\)
⇒ \(x\,dx = t\,dt\)
\(\therefore \int \frac{x - 1}{\sqrt{x^2 + 1}}dx = \int \frac{t\,dt}t - log\left|x + \sqrt{x^2 +1}\right|+ C\)
\(= t - log \left|x + \sqrt{x^2 + 1}\right| + C\)
\(= \sqrt{x^2 + 1} - log \left|x + \sqrt{x^2 + 1}\right| + C\)
