Answer
$2^{3/2}-1$
Work Step by Step
Here, we have
$ ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
and $ds= 2 \sqrt {t^2+1} dt$
This implies that
$L=\int_{0}^{1} (\dfrac{3}{2}) 2t \sqrt {t^2+1} dt$
Plug $t^2+1 =p \implies 2t dt =dp$
or, $ (\dfrac{3}{2}) \int_{1}^{2} p^{1/2} dp= \dfrac{3}{2} [p^{3/2}]_{1}^{2}=2^{3/2}-1$
