Question

Evaluate the integral: \[ I = \int \frac{1}{x^2\sqrt{1 + x^2}} \,dx. \]

• A. \( \frac{-\sqrt{x^2 + 1}}{x} + c \)
• B. \( \frac{\sqrt{x^2 + 1}}{x} + c \)
• C. \( \frac{-\sqrt{x^2 - 1}}{x} + c \)
• D. \( \frac{\sqrt{x^2 - 1}}{x} + c \)

Hint:

For integrals involving square roots of quadratic expressions, look for substitutions like \( u = \sqrt{1 + x^2} \) to simplify the integral.

Answer 1

The Correct Option is A

Step 1: Use substitution Let \[ I = \int \frac{1}{x^2\sqrt{1 + x^2}} \,dx. \] We use the substitution: \[ u = \sqrt{1 + x^2}. \] Differentiating both sides: \[ du = \frac{x}{\sqrt{1 + x^2}} dx. \] Rearranging: \[ du \cdot \sqrt{1 + x^2} = x \,dx. \]
Step 2: Expressing in terms of \( u \) Rewriting the integral: \[ I = \int \frac{du}{x^2}. \] Since \( x^2 = u^2 - 1 \), \[ I = \int \frac{du}{(u^2 - 1)}. \] Recognizing the standard form of integration, \[ I = -\frac{\sqrt{x^2 + 1}}{x} + c. \] % Final Answer Thus, the correct answer is option (1).