Answer
$$L = 2\left( {{e^4} - e} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}x = {e^t}\left( {\sin t + \cos t} \right),{\text{ }}y = {e^t}\left( {\cos t - \sin t} \right),{\text{ }}\left( {1 \leqslant t \leqslant 4} \right) \cr
& {\text{Find the arc length using }}L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} } dt \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{e^t}\cos t} \right] = 2{e^t}\cos t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{e^t}\sin t} \right] = - 2{e^t}\sin t \cr
& L = \int_1^4 {\sqrt {{{\left( {2{e^t}\cos t} \right)}^2} + {{\left( { - 2{e^t}\sin t} \right)}^2}} dt} \cr
& {\text{Simplifying}} \cr
& L = \int_1^4 {\sqrt {4{e^{2t}}{{\cos }^2}t + 4{e^{2t}}{{\sin }^2}t} dt} \cr
& L = \int_1^4 {\sqrt {4{e^{2t}}\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} dt} \cr
& L = \int_1^4 {\sqrt {4{e^{2t}}} dt} \cr
& L = \int_1^4 {2{e^t}dt} \cr
& {\text{Integrating}} \cr
& L = 2\left[ {{e^t}} \right]_1^4 \cr
& L = 2\left( {{e^4} - e} \right) \cr} $$
