\(\int_{1}^{\sqrt{3}}\displaystyle\frac{1}{1+x^2}\,dx\) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{\pi}{12}\)
The standard integral \(\int \frac{dx}{1 + x^2} = \tan^{-1}x + C\) gives
\( \displaystyle \int_{1}^{\sqrt{3}} \frac{1}{1 + x^2}\,dx = \Big[\tan^{-1}x\Big]_{1}^{\sqrt{3}} \)
\( \displaystyle = \tan^{-1}\sqrt{3} - \tan^{-1}1 \)
\( \displaystyle = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} \)