Using the substitution \(x = \sqrt{3} \tan \theta\), we have \(dx = \sqrt{3} \sec^2 \theta \, d\theta\).

The integral becomes:

\(\int \frac{1}{\sqrt{3 + 3 \tan^2 \theta}} \cdot \sqrt{3} \sec^2 \theta \, d\theta = \int \sec \theta \, d\theta.\)

We need to change the limits of integration. When \(x = 1\), \(\sqrt{3} \tan \theta = 1\), so \(\tan \theta = \frac{1}{\sqrt{3}}\), giving \(\theta = \frac{\pi}{6}\).

When \(x = 3\), \(\sqrt{3} \tan \theta = 3\), so \(\tan \theta = \sqrt{3}\), giving \(\theta = \frac{\pi}{3}\).

Thus, the integral becomes:

\(\int_{\pi/6}^{\pi/3} \sec \theta \, d\theta.\)

The integral of \(\sec \theta\) is \(\ln |\sec \theta + \tan \theta|\).

Evaluating from \(\theta = \frac{\pi}{6}\) to \(\theta = \frac{\pi}{3}\):

\(\ln |\sec \frac{\pi}{3} + \tan \frac{\pi}{3}| - \ln |\sec \frac{\pi}{6} + \tan \frac{\pi}{6}|.\)

\(\sec \frac{\pi}{3} = 2\), \(\tan \frac{\pi}{3} = \sqrt{3}\), \(\sec \frac{\pi}{6} = \frac{2}{\sqrt{3}}\), \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\).

Thus, the expression becomes:

\(\ln (2 + \sqrt{3}) - \ln \left( \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right) = \ln (2 + \sqrt{3}) - \ln \left( \frac{3}{\sqrt{3}} \right).\)

\(= \ln (2 + \sqrt{3}) - \ln \sqrt{3} = \ln \left( \frac{2 + \sqrt{3}}{\sqrt{3}} \right).\)