(i) Let \(x = \sqrt{3} \tan \theta\). Then \(dx = \sqrt{3} \sec^2 \theta \, d\theta\).

Substitute into the integral:

\(I = \int_0^1 \frac{9}{(3 + x^2)^2} \, dx = \int_0^{\frac{\pi}{6}} \frac{9}{(3 + 3 \tan^2 \theta)^2} \cdot \sqrt{3} \sec^2 \theta \, d\theta\).

\(= \sqrt{3} \int_0^{\frac{\pi}{6}} \frac{9 \sec^2 \theta}{(3 \sec^2 \theta)^2} \, d\theta\).

\(= \sqrt{3} \int_0^{\frac{\pi}{6}} \cos^2 \theta \, d\theta\).

(ii) Use the identity \(\cos^2 \theta = \frac{1}{2} \cos 2\theta + \frac{1}{2}\).

\(I = \sqrt{3} \int_0^{\frac{\pi}{6}} \left( \frac{1}{2} \cos 2\theta + \frac{1}{2} \right) \, d\theta\).

\(= \sqrt{3} \left[ \frac{1}{4} \sin 2\theta + \frac{1}{2} \theta \right]_0^{\frac{\pi}{6}}\).

\(= \sqrt{3} \left( \frac{1}{4} \sin \frac{\pi}{3} + \frac{1}{2} \cdot \frac{\pi}{6} \right)\).

\(= \sqrt{3} \left( \frac{1}{4} \cdot \frac{\sqrt{3}}{2} + \frac{\pi}{12} \right)\).

\(= \frac{1}{12} \sqrt{3} \pi + \frac{3}{8}\).