Answer
$\frac{1}{2} \sinh 2$
Work Step by Step
Given: $y = 3+ \frac{1}{2} \cosh 2x$
$y' = \sinh 2x$
and $1+(dy/dx)^{2} = 1 + \sinh^{2} 2x = \cosh^{2} 2x$
$L = \int^{1}_{0} \sqrt{\cosh^{2} 2x} dx $
$= \int^{1}_{0} \cosh 2x dx $
$= [\frac{1}{2} \sinh 2x]^{1}_{0}$
$ = \frac{1}{2} \sinh 2 - 0 $
$=\frac{1}{2} \sinh 2$