(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 \(\cos^2 \theta = \frac{1}{2} \cos 2\theta + \frac{1}{2}\).

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

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

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

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

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

Let \(u = 4 - 3 \cos x\). Then, \(du = 3 \sin x \, dx\).

Using the identity \(\sin 2x = 2 \sin x \cos x\), we have:

\(\int \frac{9 \sin 2x}{\sqrt{4 - 3 \cos x}} \, dx = \int \frac{18 \sin x \cos x}{\sqrt{u}} \, dx\).

Substitute \(dx = \frac{du}{3 \sin x}\), so:

\(\int \frac{18 \sin x \cos x}{\sqrt{u}} \cdot \frac{du}{3 \sin x} = \int \frac{6 \cos x}{\sqrt{u}} \, du\).

Since \(\cos x = \frac{4 - u}{3}\), the integral becomes:

\(\int \frac{6 \left(\frac{4 - u}{3}\right)}{\sqrt{u}} \, du = \int \frac{8 - 2u}{\sqrt{u}} \, du\).

Integrate to obtain:

\(\int \frac{8}{\sqrt{u}} \, du - \int \frac{2u}{\sqrt{u}} \, du = 16u^{\frac{1}{2}} - \frac{4}{3}u^{\frac{3}{2}}\).

Apply the limits from \(x = 0\) to \(x = \frac{1}{2}\pi\):

When \(x = 0, u = 4 - 3 \cdot 1 = 1\).

When \(x = \frac{1}{2}\pi, u = 4 - 3 \cdot 0 = 4\).

Evaluate:

\(\left[ 16u^{\frac{1}{2}} - \frac{4}{3}u^{\frac{3}{2}} \right]_1^4 = \left( 16 \cdot 2 - \frac{4}{3} \cdot 8 \right) - \left( 16 \cdot 1 - \frac{4}{3} \cdot 1 \right)\).

\(= (32 - \frac{32}{3}) - (16 - \frac{4}{3})\).

\(= \frac{64}{3} - \frac{44}{3} = \frac{20}{3}\).

(i) Let \(u = 2 - \sqrt{x}\). Then \(\sqrt{x} = 2 - u\) and \(x = (2-u)^2\).

Differentiating, \(\frac{du}{dx} = -\frac{1}{2\sqrt{x}}\), so \(du = -\frac{1}{2\sqrt{x}} \, dx\).

Thus, \(dx = -2\sqrt{x} \, du = -2(2-u) \, du\).

Substitute into the integral:

\(I = \int_0^1 \frac{\sqrt{x}}{2 - \sqrt{x}} \, dx = \int_1^2 \frac{2(2-u)^2}{u} \, du\).

(ii) Integrate \(\int_1^2 \frac{2(2-u)^2}{u} \, du\):

\(\int \frac{2(2-u)^2}{u} \, du = \int \left( \frac{8}{u} - 8 + u \right) \, du\).

\(= 8 \ln u - 8u + \frac{u^2}{2} + C\).

Evaluate from 1 to 2:

\(\left[ 8 \ln u - 8u + \frac{u^2}{2} \right]_1^2 = \left( 8 \ln 2 - 16 + 2 \right) - \left( 0 - 8 + \frac{1}{2} \right)\).

\(= 8 \ln 2 - 14 + 8 - \frac{1}{2}\).

\(= 8 \ln 2 - 5\).