(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:

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

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

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