Answer
$$S = 2\pi \left( {\frac{{195}}{{128}} + \ln \sqrt 2 } \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = {e^x} + \frac{1}{4}{e^{ - x}}{\text{ on the interval }}\left[ {0,\ln 2} \right]{\text{ is revolved about }}x \cr
& f\left( x \right) = {e^x} + \frac{1}{4}{e^{ - x}} \cr
& {\text{The area of surface is given by}} \cr
& S = \int_a^b {2\pi f\left( x \right)\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^x} + \frac{1}{4}{e^{ - x}}} \right] \cr
& f'\left( x \right) = {e^x} - \frac{1}{4}{e^{ - x}} \cr
& S = 2\pi \int_0^{\ln 2} {\left( {{e^x} + \frac{1}{4}{e^{ - x}}} \right)\sqrt {1 + {{\left( {{e^x} - \frac{1}{4}{e^{ - x}}} \right)}^2}} dx} \cr
& S = 2\pi \int_0^{\ln 2} {\left( {{e^x} + \frac{1}{4}{e^{ - x}}} \right)\sqrt {1 + {e^{2x}} - \frac{1}{2}{e^0} + \frac{1}{{16}}{e^{ - 2x}}} dx} \cr
& S = 2\pi \int_0^{\ln 2} {\left( {{e^x} + \frac{1}{4}{e^{ - x}}} \right)\sqrt {{e^{2x}} + \frac{1}{2} + \frac{1}{{16}}{e^{ - 2x}}} dx} \cr
& {\text{Factoring}} \cr
& S = 2\pi \int_0^{\ln 2} {\left( {{e^x} + \frac{1}{4}{e^{ - x}}} \right)\sqrt {{{\left( {{e^x} + \frac{1}{4}{e^{ - x}}} \right)}^2}} dx} \cr
& S = 2\pi \int_0^{\ln 2} {\left( {{e^x} + \frac{1}{4}{e^{ - x}}} \right)\left( {{e^x} + \frac{1}{4}{e^{ - x}}} \right)dx} \cr
& S = 2\pi \int_0^{\ln 2} {{{\left( {{e^x} + \frac{1}{4}{e^{ - x}}} \right)}^2}dx} \cr
& {\text{Expanding the binomial}} \cr
& S = 2\pi \int_0^{\ln 2} {\left( {{e^{2x}} + 2{e^x}\left( {\frac{1}{4}{e^{ - x}}} \right) + \frac{1}{{16}}{e^{ - 2x}}} \right)dx} \cr
& S = 2\pi \int_0^{\ln 2} {\left( {{e^{2x}} + \frac{1}{2} + \frac{1}{{16}}{e^{ - 2x}}} \right)dx} \cr
& {\text{Integrating}} \cr
& S = 2\pi \left[ {\frac{1}{2}{e^{2x}} + \frac{1}{2}x - \frac{1}{{32}}{e^{ - 2x}}} \right]_0^{\ln 2} \cr
& S = 2\pi \left[ {\frac{1}{2}{e^{2\ln 2}} + \frac{1}{2}\ln 2 - \frac{1}{{32}}{e^{ - 2\ln 2}}} \right] - 2\pi \left[ {\frac{1}{2}{e^0} + \frac{1}{2}\left( 0 \right) - \frac{1}{{32}}{e^0}} \right] \cr
& S = 2\pi \left[ {2 + \ln \sqrt 2 - \frac{1}{{128}}} \right] - 2\pi \left[ {\frac{1}{2} - \frac{1}{{32}}} \right] \cr
& S = 2\pi \left( {\frac{{255}}{{128}} + \ln \sqrt 2 - \frac{{15}}{{32}}} \right) \cr
& S = 2\pi \left( {\frac{{195}}{{128}} + \ln \sqrt 2 } \right) \cr} $$
