1. **State the problem:** Find the Laplace transform of the function $$f(t) = e^{2t} (\sinh(t) + \cosh(t))$$. 2. **Recall the definitions and formulas:** - The Laplace transform of a function $$f(t)$$ is defined as $$\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt$$. - Important hyperbolic identities: $$\sinh(t) = \frac{e^t - e^{-t}}{2}$$ $$\cosh(t) = \frac{e^t + e^{-t}}{2}$$ - Sum of hyperbolic sine and cosine: $$\sinh(t) + \cosh(t) = \frac{e^t - e^{-t}}{2} + \frac{e^t + e^{-t}}{2} = e^t$$ 3. **Simplify the function inside the Laplace transform:** $$f(t) = e^{2t} (\sinh(t) + \cosh(t)) = e^{2t} \cdot e^t = e^{3t}$$ 4. **Apply the Laplace transform formula:** $$\mathcal{L}\{e^{3t}\} = \int_0^\infty e^{-st} e^{3t} dt = \int_0^\infty e^{(3 - s)t} dt$$ 5. **Evaluate the integral:** For convergence, we require $$s > 3$$. $$\int_0^\infty e^{(3 - s)t} dt = \left[ \frac{e^{(3 - s)t}}{3 - s} \right]_0^\infty = \lim_{T \to \infty} \frac{e^{(3 - s)T} - 1}{3 - s}$$ Since $$s > 3$$, $$3 - s < 0$$, so $$e^{(3 - s)T} \to 0$$ as $$T \to \infty$$. Therefore, $$\mathcal{L}\{e^{3t}\} = \frac{0 - 1}{3 - s} = \frac{1}{s - 3}$$ **Final answer:** $$\boxed{\mathcal{L}\{e^{2t} (\sinh(t) + \cosh(t))\} = \frac{1}{s - 3}, \quad s > 3}$$