Answer
$$s = \int_1^3 {\sqrt {1 + \frac{1}{{{x^4}}}} } dx \approx 2.14662$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{x},{\text{ }}1 \leqslant x \leqslant 3 \cr
& \cr
& \left( {\text{a}} \right){\text{Graph below}} \cr
& \cr
& {\text{Use the formula for arc length}} \cr
& s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{1}{x}} \right] \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{x^2}}} \cr
& s = \int_1^3 {\sqrt {1 + {{\left( { - \frac{1}{{{x^2}}}} \right)}^2}} } dx \cr
& s = \int_1^3 {\sqrt {1 + \frac{1}{{{x^4}}}} } dx \cr
& \cr
& \left( {\text{c}} \right){\text{Integrate by using a CAS or graphing utility}} \cr
& s = \int_1^3 {\sqrt {1 + \frac{1}{{{x^4}}}} } dx \approx 2.14662 \cr
& \cr
& {\text{Graph}} \cr} $$
