To expand \((2 + x^2)^{-2}\), we use the binomial series expansion for negative exponents:

\((1 + u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 - \frac{n(n+1)(n+2)}{3!}u^3 + \cdots\)

Here, \(u = \frac{x^2}{2}\) and \(n = 2\).

First term: \(\frac{1}{4}\).

Second term: \(-2 \times \frac{x^2}{2} = -x^2\).

Third term: \(\frac{2 \times 3}{2} \left(\frac{x^2}{2}\right)^2 = \frac{3}{16}x^4\).

Thus, the expansion up to \(x^4\) is \(\frac{1}{4} - \frac{1}{4}x^2 + \frac{3}{16}x^4\).