Question

Evaluate \[ \int \frac{x^2}{x^3\sqrt{x^2-1}}\,dx = \]

• A. \[ \frac{-x^2}{\sqrt{x^2-1}} \]
• B. \[ \frac{-\sqrt{x^2-1}}{x} \]
• C. \[ \frac{-x}{\sqrt{x^2-1}} \]
• D. \[ \frac{-\sqrt{x^2-1}}{x^2} \]

Hint:

For integrals involving \[ \sqrt{x^2-a^2}, \] try identifying derivatives of expressions like \[ \frac{\sqrt{x^2-a^2}}{x} \] before using substitutions.

Answer 1

The Correct Option is D

Step 1: Simplify the integrand.
Given, \[ \int \frac{x^2}{x^3\sqrt{x^2-1}}\,dx \] Cancel \(x^2\) from numerator and denominator: \[ = \int \frac{1}{x\sqrt{x^2-1}}\,dx \]

Step 2: Use standard differentiation.
Consider \[ f(x)=\frac{\sqrt{x^2-1}}{x} \] Differentiate using quotient rule: \[ f'(x) = \frac{ x\left(\frac{x}{\sqrt{x^2-1}}\right)-\sqrt{x^2-1} }{ x^2 } \] \[ = \frac{ \frac{x^2}{\sqrt{x^2-1}}-\sqrt{x^2-1} }{ x^2 } \] Take LCM in numerator: \[ = \frac{ x^2-(x^2-1) }{ x^2\sqrt{x^2-1} } \] \[ = \frac{1}{x^2\sqrt{x^2-1}} \] Therefore, \[ \frac{d}{dx} \left( -\frac{\sqrt{x^2-1}}{x} \right) = -\frac{1}{x^2\sqrt{x^2-1}} \] Hence, \[ \int \frac{1}{x^2\sqrt{x^2-1}}\,dx = -\frac{\sqrt{x^2-1}}{x}+C \]

Step 3: Match with the given options.
Thus, \[ \boxed{ -\frac{\sqrt{x^2-1}}{x} } \] which corresponds to option (2).

Step 4: Final conclusion.
Hence, \[ \boxed{ -\frac{\sqrt{x^2-1}}{x} } \]