To solve \(\int_{1}^{\infty} (4x + 2)^{-2} \, dx\), we first find the antiderivative of \((4x + 2)^{-2}\).

Let \(u = 4x + 2\), then \(du = 4 \, dx\) or \(dx = \frac{1}{4} \, du\).

The integral becomes \(\int (4x + 2)^{-2} \, dx = \int u^{-2} \cdot \frac{1}{4} \, du\).

This simplifies to \(\frac{1}{4} \int u^{-2} \, du\).

The antiderivative of \(u^{-2}\) is \(-u^{-1}\), so we have:

\(\frac{1}{4} \cdot (-u^{-1}) = -\frac{1}{4u}\).

Substitute back \(u = 4x + 2\):

\(-\frac{1}{4(4x + 2)} = -\frac{1}{16x + 8}\).

Now, evaluate the definite integral from 1 to \(\infty\):

\(\left[ -\frac{1}{16x + 8} \right]_{1}^{\infty} = \left( 0 - \left(-\frac{1}{24}\right) \right) = \frac{1}{24}\).