Question

Evaluate: \[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx \]

Hint:

This is a standard integral result. Always recall: \[ \int \frac{1}{\sqrt{x^2 - a^2}}dx = \ln|x + \sqrt{x^2 - a^2}| + C \] It comes from trigonometric substitution $x = a \sec \theta$.

Answer 1

Step 1: Standard formula. We know, \[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln \left| x + \sqrt{x^2 - a^2} \right| + C, (|x| > a) \]

Step 2: Apply directly. \[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln \left| x + \sqrt{x^2 - a^2} \right| + C \]

Final Answer: \[ \boxed{\ln \left| x + \sqrt{x^2 - a^2} \right| + C} \]