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

Substitute into the integral:

\(\int \frac{1-x^2}{(1+x^2)^2} \, dx = \int \frac{1-\tan^2 \theta}{(1+\tan^2 \theta)^2} \sec^2 \theta \, d\theta.\)

Using the identity \(1 + \tan^2 \theta = \sec^2 \theta\), the integral becomes:

\(\int \frac{1-\tan^2 \theta}{\sec^4 \theta} \sec^2 \theta \, d\theta = \int \frac{\sec^2 \theta - \tan^2 \theta}{\sec^4 \theta} \sec^2 \theta \, d\theta.\)

Simplify to:

\(\int \frac{\sec^2 \theta - \tan^2 \theta}{\sec^2 \theta} \, d\theta = \int (1 - \sin^2 \theta) \, d\theta = \int \cos 2\theta \, d\theta.\)

(ii) From part (i), we have:

\(\int_0^1 \frac{1-x^2}{(1+x^2)^2} \, dx = \int_0^{\frac{\pi}{4}} \cos 2\theta \, d\theta.\)

The integral of \(\cos 2\theta\) is \(\frac{1}{2} \sin 2\theta\).

Evaluate from \(\theta = 0\) to \(\theta = \frac{\pi}{4}\):

\(\frac{1}{2} \sin 2\theta \bigg|_0^{\frac{\pi}{4}} = \frac{1}{2} (\sin \frac{\pi}{2} - \sin 0) = \frac{1}{2} (1 - 0) = \frac{1}{2}.\)