For the given curve, the line element is
[tex]ds = \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2} \, dt = \sqrt{9t^4 + 16t^6} \, dt[/tex]
since
[tex]x = t^3 \implies \dfrac{dx}{dt} = 3t^2[/tex]
[tex]y = t^4 \implies \dfrac{dy}{dt} = 4t^3[/tex]
Then the line integral is
[tex]\displaystyle \int_C \frac xy \, ds = \int_1^2 \frac{t^3}{t^4} \, ds = \int_1^2 \frac{\sqrt{9t^4+16t^6}}t \, dt[/tex]
Simplify the integrand to
[tex]\displaystyle \int_1^2 t \sqrt{9 + 16t^2} \, dt[/tex]
and substitute
[tex]u = 9 + 16t^2 \implies du = 32t \, dt[/tex]
Then the integral is
[tex]\displaystyle \int_1^2 t \sqrt{9+16t^2} \, dt = \frac1{32} \int_{25}^{73} \sqrt{u} \, du = \frac{73^{\frac32} - 25^{\frac32}}{48} = \boxed{\frac{73\sqrt{73} - 125}{48}}[/tex]
