$\int_1^{\sqrt{3}} \frac{1}{1+x^2} dx$ is equal to:

The integral of $\frac{1}{1+x^2}$ is a standard form:

$\int \frac{1}{1+x^2} dx = \tan^{-1}(x) + C$

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Using the fundamental theorem of calculus:

$\int_1^{\sqrt{3}} \frac{1}{1+x^2} dx = \left[\tan^{-1}(x)\right]_1^{\sqrt{3}}$

$= \tan^{-1}(\sqrt{3}) - \tan^{-1}(1)$

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The angle with $\tan(\theta) = \sqrt{3}$ is $\theta = \frac{\pi}{3}$

Therefore $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$

The angle with $\tan(\theta) = 1$ is $\theta = \frac{\pi}{4}$

Therefore $\tan^{-1}(1) = \frac{\pi}{4}$

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$\tan^{-1}(\sqrt{3}) - \tan^{-1}(1) = \frac{\pi}{3} - \frac{\pi}{4}$

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

$= \frac{\pi}{12}$

Therefore, the answer is $\frac{\pi}{12}$.