The length of the curve will be given by the definite integral
[tex]\displaystyle \int_0^1 \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt[/tex]
From the given parametric equations, we get derivatives
[tex]x(t) = 3t^2 + 5 \implies \dfrac{dx}{dt} = 6t[/tex]
[tex]y(t) = 2t^3 + 5 \implies \dfrac{dy}{dt} = 6t^2[/tex]
Then the arc length integral becomes
[tex]\displaystyle \int_0^1 \sqrt{\left(6t\right)^2 + \left(6t^2\right)^2} \, dt = \int_0^1 \sqrt{36t^2 + 36t^4} \, dt \\\\ = \int_0^1 6|t| \sqrt{1 + t^2} \, dt[/tex]
Since 0 ≤ t ≤ 1, we have |t| = t, so
[tex]\displaystyle \int_0^1 6|t| \sqrt{1 + t^2} \, dt = 6 \int_0^1 t \sqrt{1 + t^2} \, dt[/tex]
For the remaining integral, substitute [tex]u = 1 + t^2[/tex] and [tex]du = 2t \, dt[/tex]. Then
[tex]\displaystyle 6 \int_0^1 t \sqrt{1 + t^2} \, dt = 3 \int_1^2 \sqrt{u} \, du \\\\ = 3\times \frac23 u^{3/2} \bigg|_{u=1}^2 \\\\ = 2 \left(2^{3/2} - 1^{3/2}\right) = 2^{5/2} - 2 = \boxed{4\sqrt2-2}[/tex]
