Question

\[\int\frac{x - 1}{\sqrt{x^2 + 1}} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\left( \frac{x - 1}{\sqrt{x^2 + 1}} \right) dx\]
`  =  ∫   {x    dx}/\sqrt{x^2 + 1} -   ∫   dx / \sqrt{x^2 +1 }`
\[\text{ Putting x}^2 + 1 = t\]
\[ \Rightarrow \text{ 2x dx } = dt\]
\[ \Rightarrow \text{ x dx } = \frac{dt}{2}\]
\[\text{ Then }, \]
\[I = \frac{1}{2}\int\frac{dt}{\sqrt{t}} - \int\frac{dx}{\sqrt{x^2 + 1^2}}\]
\[ = \frac{1}{2}\int t^{- \frac{1}{2}} dt - \int\frac{dx}{\sqrt{x^2 + 1^2}}\]
\[ = \frac{1}{2} \left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] - \int\frac{dx}{\sqrt{x^2 + 1^2}}\]
\[ = \sqrt{t} - \text{ log }\left| x + \sqrt{x^2 + 1} \right| + C\]
\[ = \sqrt{x^2 + 1} - \text{ log }\left| x + \sqrt{x^2 + 1} \right| + C\]