Getting tired of these yet?

We have $x(t)=\frac{20}{29}\sinh^{-1} t-\frac{21}{29} \sqrt{1+t^2} \implies x'(t)=\frac{20-21t}{29\sqrt{1+t^2}}$ $y(t)=\frac{20}{29}\sqrt{1+t^2}+\frac{21}{29}\sinh^{-1} t \implies y'(t)=\frac{21+20t}{29\sqrt{1+t^2}}.$ So, the arc length is given by $\int_{-10}^{10} \sqrt{[x'(t)]^2+[y'(t)]^2} \, dt = \int_{-10}^{10} \sqrt{\left(\frac{20-21t}{29\sqrt{1+t^2}}\right)^2+\left(\frac{21+20t}{29\sqrt{1+t^2}}\right)^2}=\int_{-10}^{10} \sqrt{\frac{20^2+21^2}{29^2}} \, dt=\int_{-10}^{10} 1 \,dt=\boxed{20}.$