University Calculus: Early Transcendentals (3rd Edition)

$4\sqrt 3$
Since, $L=\int_{-\sqrt 3}^{\sqrt 3}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}dt$ Thus, $L=\int_{-\sqrt 3}^{\sqrt 3} \sqrt{(t^2+1)^2} dt$ Then, we have $L=[\dfrac{t^3}{3}+t]_{-\sqrt 3}^{\sqrt 3}$ Thus, $L =4\sqrt 3$