To solve the given integral using the suggested trick, we want to evaluate:

$I = \int_0^\infty \frac{e^{-2x} - e^{-4x}}{x} \, dx$

Since the integral doesn't have a straightforward anti-derivative, we will use the double integral method by finding a function $f(x, y)$ such that:

$\int_a^b f(x, y) \, dy = \frac{e^{-2x} - e^{-4x}}{x}$

Step 1: Setting up the double integral representation

We start by expressing the numerator in terms of integrals involving $y$:

$e^{-2x} = \int_0^\infty e^{-yt} \delta(y - 2x) \, dy \quad \text{and} \quad e^{-4x} = \int_0^\infty e^{-yt} \delta(y - 4x) \, dy$

where $\delta(y)$ is the Dirac delta function. These representations will help simplify the manipulation later.

Step 2: Using a Laplace transform-like representation

A useful trick here is to express $e^{-2x}$ and $e^{-4x}$ using integrals over auxiliary variables. We can write each of these terms as follows:

[ e^{-2x} = \int_0^\infty e^{-yt} , dy \quad \text{and} \quad e^{-4x} = \int_0^\infty e^{-yt} , dy