1. **State the problem:** Calculate the integral $$\int_0^{\ln 2} 2\pi \left( \frac{e^y + e^{-y}}{2} \right) \sqrt{1 + \left( \frac{e^y - e^{-y}}{2} \right)^2} \, dy$$. 2. **Rewrite the integral using hyperbolic functions:** Recall that $$\cosh y = \frac{e^y + e^{-y}}{2}$$ and $$\sinh y = \frac{e^y - e^{-y}}{2}$$. So the integral becomes: $$\int_0^{\ln 2} 2\pi \cosh y \sqrt{1 + \sinh^2 y} \, dy$$. 3. **Simplify the square root:** Using the identity $$1 + \sinh^2 y = \cosh^2 y$$, we get: $$\sqrt{1 + \sinh^2 y} = \sqrt{\cosh^2 y} = |\cosh y|$$. Since $$\cosh y > 0$$ for all real $$y$$, this simplifies to $$\cosh y$$. 4. **Substitute back into the integral:** $$\int_0^{\ln 2} 2\pi \cosh y \cdot \cosh y \, dy = \int_0^{\ln 2} 2\pi \cosh^2 y \, dy$$. 5. **Use the identity for $$\cosh^2 y$$:** $$\cosh^2 y = \frac{1 + \cosh 2y}{2}$$. 6. **Rewrite the integral:** $$\int_0^{\ln 2} 2\pi \cdot \frac{1 + \cosh 2y}{2} \, dy = \pi \int_0^{\ln 2} (1 + \cosh 2y) \, dy$$. 7. **Integrate term by term:** $$\int_0^{\ln 2} 1 \, dy = y \Big|_0^{\ln 2} = \ln 2$$. $$\int_0^{\ln 2} \cosh 2y \, dy = \frac{\sinh 2y}{2} \Big|_0^{\ln 2}$$. 8. **Evaluate $$\sinh 2y$$ at the bounds:** $$\sinh 2(\ln 2) = \sinh (\ln 4) = \frac{e^{\ln 4} - e^{-\ln 4}}{2} = \frac{4 - \frac{1}{4}}{2} = \frac{\frac{16}{4} - \frac{1}{4}}{2} = \frac{15/4}{2} = \frac{15}{8}$$. At $$y=0$$, $$\sinh 0 = 0$$. 9. **Substitute back:** $$\int_0^{\ln 2} \cosh 2y \, dy = \frac{15}{8} - 0 = \frac{15}{8}$$. 10. **Combine results:** $$\pi \left( \ln 2 + \frac{15}{8} \right) = \pi \ln 2 + \frac{15\pi}{8}$$. **Final answer:** $$\boxed{\pi \ln 2 + \frac{15\pi}{8}}$$