Question

(c) The value of the integral: \[ \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{dx}{1+x^2} \]

• A. \(\frac{\pi}{3}\)
• B. \(\frac{2\pi}{3}\)
• C. \(\frac{\pi}{6}\)
• D. \(\frac{\pi}{12}\)

Hint:

Use standard trigonometric integrals and apply limits step by step.

Answer 1

The Correct Option is A

The given integral is a standard form: \[ \int \frac{dx}{1+x^2} = \tan^{-1}(x). \] Applying limits: \[ \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{dx}{1+x^2} = \tan^{-1}(\sqrt{3}) - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right). \] \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}, \quad \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}. \] \[ \text{Result: } \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}. \]