Question

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

Answer 1

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