[tex]x=1+6t^2\implies\dfrac{\mathrm dx}{\mathrm dt}=12t[/tex]
[tex]y=4+4t^3\implies\dfrac{\mathrm dy}{\mathrm dt}=12t^2[/tex]
The length of the curve [tex]C[/tex] is given by
[tex]\displaystyle\int_C\mathrm ds=\int_0^3\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^3\sqrt{144t^2+144t^4}\,\mathrm dt[/tex]
[tex]=\displaystyle12\int_0^3t\sqrt{1+t^2}\,\mathrm dt[/tex]
[tex]=\displaystyle6\int_0^3 2t\sqrt{1+t^2}\,\mathrm dt[/tex]
[tex]=\displaystyle6\int_0^3\sqrt{1+t^2}\,\mathrm d(1+t^2)[/tex]
[tex]=6\cdot\dfrac23(1+t^2)^{3/2}\bigg|_0^3[/tex]
[tex]=4(10^{3/2}-1)[/tex]
[tex]=40\sqrt{10}-4[/tex]
