Let \(u = \sqrt{x}\), then \(x = u^2\) and \(dx = 2u \, du\).

The limits change from \(x = 3\) to \(u = \sqrt{3}\) and \(x = \infty\) to \(u = \infty\).

Substitute into the integral:

\(\int_{\sqrt{3}}^{\infty} \frac{1}{(u^2 + 1)u} \, 2u \, du = \int_{\sqrt{3}}^{\infty} \frac{2}{u^2 + 1} \, du\).

The integral \(\int \frac{2}{u^2 + 1} \, du\) is \(2 \arctan(u)\).

Evaluate from \(\sqrt{3}\) to \(\infty\):

\(2 \left[ \arctan(\infty) - \arctan(\sqrt{3}) \right] = 2 \left[ \frac{\pi}{2} - \frac{\pi}{3} \right]\).

\(= 2 \left[ \frac{\pi}{6} \right] = \frac{1}{3} \pi\).