(i) Let \(x = \sin^2 \theta\). Then \(dx = 2 \sin \theta \cos \theta \, d\theta\).

Substitute into the integral:

\(\int \sqrt{\left( \frac{x}{1-x} \right)} \, dx = \int \sqrt{\left( \frac{\sin^2 \theta}{1-\sin^2 \theta} \right)} \, 2 \sin \theta \cos \theta \, d\theta\).

Since \(1 - \sin^2 \theta = \cos^2 \theta\), the expression simplifies to:

\(\int \sqrt{\left( \frac{\sin^2 \theta}{\cos^2 \theta} \right)} \, 2 \sin \theta \cos \theta \, d\theta = \int 2 \sin^2 \theta \, d\theta\).

(ii) Use the result from part (i):

\(\int_0^{\frac{1}{4}} \sqrt{\left( \frac{x}{1-x} \right)} \, dx = \int_0^{\frac{\pi}{6}} 2 \sin^2 \theta \, d\theta\).

Using the identity \(\sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta)\), the integral becomes:

\(\int_0^{\frac{\pi}{6}} (1 - \cos 2\theta) \, d\theta = \left[ \theta - \frac{1}{2} \sin 2\theta \right]_0^{\frac{\pi}{6}}\).

Evaluating the limits:

\(\left( \frac{\pi}{6} - \frac{1}{2} \sin \frac{\pi}{3} \right) - \left( 0 - 0 \right) = \frac{\pi}{6} - \frac{1}{4} \sqrt{3}\).