Question

\( \int \frac{\sqrt{x^2 - 1}}{x} \, dx \) is equal to:

• A. \( \sqrt{x^2 - 1} - \sec^{-1} x + C \)
• B. \( \sqrt{x^2 - 1} + \tan^{-1} x + C \)
• C. \( \sqrt{x^2 - 1} + \sec^{-1} x + C \)
• D. \( \sqrt{x^2 - 1} - \tan x + C \)
• E. \( \sqrt{x^2 - 1} + \sec x + C \)

Hint:

Trigonometric substitution turns algebraic radicals into powers of trig functions. Remember the triangle: if $\sec\theta = x/1$, the opposite side is $\sqrt{x^2-1}$.

Answer 1

The Correct Option is A

Concept: For integrals involving \( \sqrt{x^2 - a^2} \), the trigonometric substitution \( x = a\sec\theta \) is effective. Here, \( a = 1 \).

Step 1: Apply trigonometric substitution.
Let \( x = \sec\theta \). Then \( dx = \sec\theta \tan\theta \, d\theta \). Also, \( \sqrt{x^2 - 1} = \sqrt{\sec^2\theta - 1} = \tan\theta \). The integral becomes: \[ \int \frac{\tan\theta}{\sec\theta} \cdot \sec\theta \tan\theta \, d\theta = \int \tan^2\theta \, d\theta \]

Step 2: Integrate.
Using the identity \( \tan^2\theta = \sec^2\theta - 1 \): \[ \int (\sec^2\theta - 1) \, d\theta = \tan\theta - \theta + C \]

Step 3: Back-substitute.
Since \( \sec\theta = x \), we have \( \tan\theta = \sqrt{x^2 - 1} \) and \( \theta = \sec^{-1} x \). \[ \text{Result} = \sqrt{x^2 - 1} - \sec^{-1} x + C \]