Answer
$$\frac{1}{{24}}{e^{3x}} + \frac{3}{8}{e^x} - \frac{3}{8}{e^{ - x}} - \frac{1}{{24}}{e^{ - 3x}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right)}^3}} dx \cr
& {\text{Expand, recall that }}{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} \cr
& = \frac{1}{8}\int {\left( {{e^{3x}} + 3{e^{2x}}{e^{ - x}} + 3{e^x}{e^{ - 2x}} + {e^{ - 3x}}} \right)} dx \cr
& = \frac{1}{8}\int {\left( {{e^{3x}} + 3{e^x} + 3{e^{ - x}} + {e^{ - 3x}}} \right)} dx \cr
& {\text{Integrate}} \cr
& = \frac{1}{8}\left( {\frac{1}{3}{e^{3x}} + 3{e^x} - 3{e^{ - x}} - \frac{1}{3}{e^{ - 3x}}} \right) + C \cr
& = \frac{1}{{24}}{e^{3x}} + \frac{3}{8}{e^x} - \frac{3}{8}{e^{ - x}} - \frac{1}{{24}}{e^{ - 3x}} + C \cr} $$
