Answer
$$s = \int_0^1 {\sqrt {1 + \frac{4}{{{{\left( {{x^2} + 1} \right)}^2}}}} } dx \approx 1.870615$$
Work Step by Step
$$\eqalign{
& y = 2\arctan x,{\text{ }}0 \leqslant x \leqslant 1 \cr
& \cr
& \left( {\text{a}} \right){\text{Graph below}} \cr
& \cr
& {\text{Use the formula for arc length}} \cr
& s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {2\arctan x} \right] \cr
& \frac{{dy}}{{dx}} = \frac{2}{{{x^2} + 1}} \cr
& s = \int_0^1 {\sqrt {1 + {{\left( {\frac{2}{{{x^2} + 1}}} \right)}^2}} } dx \cr
& s = \int_0^1 {\sqrt {1 + \frac{4}{{{{\left( {{x^2} + 1} \right)}^2}}}} } dx \cr
& \cr
& \left( {\text{c}} \right){\text{Integrate by using a CAS or graphing utility}} \cr
& s = \int_0^1 {\sqrt {1 + \frac{4}{{{{\left( {{x^2} + 1} \right)}^2}}}} } dx \approx 1.870615 \cr
& \cr
& {\text{Graph}} \cr} $$
