$\int\limits_1^{\sqrt{3}}\frac{1}{1+x^2}dx$ is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{\pi}{12}$
Given integral:
$\int_{{1}^\sqrt{3}} \frac{1}{1+x^2} dx$
Standard formula: $\int \frac{1}{1+x^2} dx = \tan^{-1}x + C$
Evaluate definite integral:
$\int_{{1}^\sqrt{3}} \frac{1}{1+x^2} dx = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1)$
$= -\frac{\pi}{4} + \frac{\pi}{3} = \frac{\pi}{12}$