Answer
$12$
Work Step by Step
Here,we have $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$
This implies that
$S=\int_{0}^{4} \sqrt{t^2+(\sqrt{2t+1})^2} dt=\int_{0}^{4} (t+1) dt$
Therefore,
$S=[\dfrac{t^{2}}{2}+t]_{0}^{4} $
or, $S=\dfrac{16}{2}+4=8+4=12$