Question

The value of $\int_{1}^{\sqrt{3}} \dfrac{dx}{1+x^{2}}$ will be:

• A. $\dfrac{\pi}{6}$
• B. $\dfrac{\pi}{12}$
• C. $\dfrac{\pi}{3}$
• D. $\dfrac{2\pi}{3}$

Hint:

Always remember $\int \dfrac{dx}{1+x^{2}} = \tan^{-1}x + C$ and use trigonometric values of $\tan^{-1}(1)$ and $\tan^{-1}(\sqrt{3})$.

Answer 1

The Correct Option is A

Step 1: Recall the standard integral.
\[ \int \frac{dx}{1+x^{2}} = \tan^{-1}x + C \]

Step 2: Apply limits.
\[ \int_{1}^{\sqrt{3}} \frac{dx}{1+x^{2}} = \left[ \tan^{-1}x \right]_{1}^{\sqrt{3}} \] \[ = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \]

Step 3: Simplify.
\[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}, \tan^{-1}(1) = \frac{\pi}{4} \] \[ \therefore \int_{1}^{\sqrt{3}} \frac{dx}{1+x^{2}} = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \]

Step 4: Conclusion.
The correct answer is (B) $\dfrac{\pi}{12}$.