1. **State the problem:** Evaluate the integral $$\int_0^\infty \frac{e^x - e^{2x}}{x} \, dx$$ using the Laplace transform. 2. **Recall the Laplace transform formula:** The Laplace transform of a function $f(t)$ is defined as $$\mathcal{L}\{f(t)\}(s) = \int_0^\infty e^{-st} f(t) \, dt.$$ A useful integral formula related to the Laplace transform is $$\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x} \, dx = \ln\left(\frac{b}{a}\right),$$ valid for $a,b > 0$. 3. **Rewrite the integral:** Notice the integral has $e^x$ and $e^{2x}$ in the numerator, which can be rewritten as $$e^x = e^{-(-1)x}$$ and $$e^{2x} = e^{-(-2)x}.$$ So the integral becomes $$\int_0^\infty \frac{e^{-(-1)x} - e^{-(-2)x}}{x} \, dx.$$ Here, $a = -1$ and $b = -2$, but the formula requires positive $a,b$. 4. **Check convergence and validity:** The integral as given diverges because $e^x$ and $e^{2x}$ grow exponentially as $x \to \infty$. The integral $$\int_0^\infty \frac{e^x - e^{2x}}{x} \, dx$$ does not converge in the usual sense. 5. **Interpretation via Laplace transform:** The problem likely intends the integral $$\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x} \, dx$$ with $a,b > 0$. If we replace $x$ by $-x$ or consider the integral $$\int_0^\infty \frac{e^{-x} - e^{-2x}}{x} \, dx,$$ then the formula applies. 6. **Evaluate the corrected integral:** Using the formula, $$\int_0^\infty \frac{e^{-x} - e^{-2x}}{x} \, dx = \ln\left(\frac{2}{1}\right) = \ln(2).$$ **Final answer:** $$\boxed{\ln(2)}$$ This is the value of the integral $$\int_0^\infty \frac{e^{-x} - e^{-2x}}{x} \, dx,$$ which is the convergent form related to the original problem via Laplace transform methods.