Answer
$\dfrac{85e^4-25}{16}$
Work Step by Step
$I=\int_{2}^{5} te^{4/3(t-2)}\sqrt {{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}dt=\int_{2}^{5} te^{4/3(t-2)}\sqrt {1^2+{(\dfrac{4}{3})^{2}}}dt$
$=\int_{2}^{5} te^{4/3(t-2)}(\dfrac{5}{3})dt$
$=(\dfrac{5}{3})[(3t/4)e^{4/3(t-2)}-\dfrac{3}{4} \int e^{4/3(t-2)}dt]$
$=(\dfrac{5}{3})[(3t/4)e^{4/3(t-2)}-\dfrac{9}{16} e^{4/3(t-2)}dt+K]$
$=(\dfrac{5}{3})[(15/4)e^{4/3(5-2)}-\dfrac{9}{16} e^{4/3(5-2)}]-(\dfrac{5}{3})[(6/4)e^{4/3(2-2)}-\dfrac{9}{16} e^{4/3(2-2)}]$
$= \dfrac{85e^4-25}{16}$
