1. **State the problem:** Evaluate the integral $$\int \ln\left(2\pi e^{y}+e^{-y}\right) \frac{\left(\frac{e^{y}+e^{-y}}{2}\right) \sqrt{1+\left(\frac{e^{y}-e^{-y}}{2}\right)^2}}{2\pi e^{y}+e^{-y}} \, dy.$$ 2. **Simplify the expressions inside the integral:** Recall the hyperbolic cosine and sine definitions: $$\cosh y = \frac{e^{y}+e^{-y}}{2}, \quad \sinh y = \frac{e^{y}-e^{-y}}{2}.$$ So the integral becomes $$\int \ln\left(2\pi e^{y}+e^{-y}\right) \frac{\cosh y \sqrt{1+\sinh^2 y}}{2\pi e^{y}+e^{-y}} \, dy.$$ 3. **Simplify the square root:** Since $$1+\sinh^2 y = \cosh^2 y,$$ we have $$\sqrt{1+\sinh^2 y} = \cosh y.$$ 4. **Rewrite the integral:** The integrand is now $$\ln\left(2\pi e^{y}+e^{-y}\right) \frac{\cosh y \cdot \cosh y}{2\pi e^{y}+e^{-y}} = \ln\left(2\pi e^{y}+e^{-y}\right) \frac{\cosh^2 y}{2\pi e^{y}+e^{-y}}.$$ 5. **Express $\cosh^2 y$ in terms of exponentials:** $$\cosh^2 y = \left(\frac{e^{y}+e^{-y}}{2}\right)^2 = \frac{e^{2y} + 2 + e^{-2y}}{4}.$$ 6. **Rewrite denominator:** $$2\pi e^{y} + e^{-y} = e^{-y} (2\pi e^{2y} + 1).$$ 7. **Rewrite the integrand:** $$\ln\left(2\pi e^{y}+e^{-y}\right) \cdot \frac{\frac{e^{2y} + 2 + e^{-2y}}{4}}{2\pi e^{y} + e^{-y}} = \ln\left(2\pi e^{y}+e^{-y}\right) \cdot \frac{e^{2y} + 2 + e^{-2y}}{4(2\pi e^{y} + e^{-y})}.$$ 8. **Substitute $x = e^{y}$, so $dy = \frac{dx}{x}$:** Rewrite terms: - Numerator: $$e^{2y} + 2 + e^{-2y} = x^{2} + 2 + \frac{1}{x^{2}}.$$ - Denominator: $$2\pi e^{y} + e^{-y} = 2\pi x + \frac{1}{x}.$$ - Logarithm: $$\ln(2\pi x + \frac{1}{x}) = \ln\left(\frac{2\pi x^{2} + 1}{x}\right) = \ln(2\pi x^{2} + 1) - \ln x.$$ 9. **Rewrite the integral in terms of $x$:** $$\int \left[\ln(2\pi x^{2} + 1) - \ln x\right] \cdot \frac{x^{2} + 2 + \frac{1}{x^{2}}}{4(2\pi x + \frac{1}{x})} \cdot \frac{dx}{x}.$$ 10. **Simplify the fraction inside the integral:** Denominator: $$4\left(2\pi x + \frac{1}{x}\right) x = 4(2\pi x^{2} + 1).$$ Numerator: $$x^{2} + 2 + \frac{1}{x^{2}} = \frac{x^{4} + 2x^{2} + 1}{x^{2}} = \frac{(x^{2} + 1)^{2}}{x^{2}}.$$ 11. **Rewrite integrand:** $$\left[\ln(2\pi x^{2} + 1) - \ln x\right] \cdot \frac{(x^{2} + 1)^{2}}{4 x^{2} (2\pi x^{2} + 1)}.$$ 12. **Split the integral:** $$\int \ln(2\pi x^{2} + 1) \frac{(x^{2} + 1)^{2}}{4 x^{2} (2\pi x^{2} + 1)} dx - \int \ln x \frac{(x^{2} + 1)^{2}}{4 x^{2} (2\pi x^{2} + 1)} dx.$$ 13. **This integral is complicated and does not simplify nicely with elementary functions.** **Final answer:** The integral simplifies to $$\int \ln\left(2\pi e^{y}+e^{-y}\right) \frac{\cosh^{2} y}{2\pi e^{y}+e^{-y}} dy,$$ which can be expressed in terms of $x = e^{y}$ as $$\int \left[\ln(2\pi x^{2} + 1) - \ln x\right] \frac{(x^{2} + 1)^{2}}{4 x^{2} (2\pi x^{2} + 1)} dx.$$ This integral requires advanced techniques or numerical methods for evaluation.