Answer
$$\frac{1}{2} + \frac{1}{4}\sinh \left( 2 \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {{{\cosh }^2}x} dx \cr
& {\text{From the hyperbolic identities cos}}{{\text{h}}^2}x = \frac{{1 + \cosh 2x}}{2} \cr
& \int_0^1 {{{\cosh }^2}x} dx = \int_0^1 {\left( {\frac{{1 + \cosh 2x}}{2}} \right)} dx \cr
& = \int_0^1 {\left( {\frac{1}{2} + \frac{1}{2}\cosh 2x} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {\frac{1}{2}x + \frac{1}{4}\sinh 2x} \right]_0^1 \cr
& {\text{Evaluating}} \cr
& {\text{ = }}\left( {\frac{1}{2}\left( 1 \right) + \frac{1}{4}\sinh 2\left( 1 \right)} \right) - \left( {\frac{1}{2}\left( 0 \right) + \frac{1}{4}\sinh 2\left( 0 \right)} \right) \cr
& = \frac{1}{2} + \frac{1}{4}\sinh \left( 2 \right) - 0 \cr
& = \frac{1}{2} + \frac{1}{4}\sinh \left( 2 \right) \cr} $$
