Answer
$ \dfrac{236\sqrt{21}}{15}$
Work Step by Step
$I=\int_{0}^{1} (2t-1) (t+5) (4t)^2 \times \sqrt {{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}dt=\int_{0}^{1} (2t-1) (t+5) (4t)^2 \times \sqrt {(2)^2+(1)^2+(4)^2}dt$
$=\int_{0}^{1} (2t-1) (t+5) (4t)^2 \times \sqrt {21}dt$
$=(\sqrt{21}) [\dfrac{32t^5}{5}+\dfrac{144t^4}{4}-\dfrac{80t^3}{3}]_0^1$
$= \dfrac{236\sqrt{21}}{15}$
