Question

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

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

Hint:

For integrals involving expressions of the form \( (1 + x^2) \) and \( \sqrt{x^2 + a} \), consider trigonometric substitutions \( x = \tan \theta \) or use identity-based simplifications.

Answer 1

The Correct Option is A

Step 1: Recognizing the Integral Form
We are given the integral: \[ I = \int \frac{1}{(1 + x^2) \sqrt{x^2 + 2}} \, dx. \] Rewriting \( 1 + x^2 \) as: \[ 1 + x^2 = \frac{x^2 + 2 - 1}{x^2 + 2}. \] Step 2: Substituting a Useful Form Let: \[ t = \sqrt{x^2 + 2}. \] Then differentiating both sides: \[ dt = \frac{x}{\sqrt{x^2 + 2}} dx. \] Rewriting the integral: \[ I = \int \frac{1}{(1 + x^2) t} dx. \] Using a trigonometric substitution: \[ x = \tan \theta, \quad dx = \sec^2 \theta d\theta. \] Step 3: Evaluating the Integral
Using standard results: \[ \int \frac{1}{(1 + x^2) \sqrt{x^2 + 2}} \, dx = -\tan^{-1} \frac{\sqrt{x^2 + 2}}{|x|} + c. \] Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{-\tan^{-1} \frac{\sqrt{x^2 + 2}}{|x|} + c}. \]