$\int\limits_1^{\sqrt{3}}\frac{1}{1+x^2}dx$ is equal to:

$\frac{\pi}{12}$

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:

$\int_{{1}^\sqrt{3}} \frac{1}{1+x^2} dx = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1)$

$= -\frac{\pi}{4} + \frac{\pi}{3} = \frac{\pi}{12}$