[tex]y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt[/tex]
By the fundamental theorem of calculus,
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}[/tex]
Now the arc length over an arbitrary interval [tex](a,b)[/tex] is
[tex]\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx[/tex]
But before we compute the integral, first we need to make sure the integrand exists over it. [tex]x^{3/2}[/tex] is undefined if [tex]x<0[/tex], so we assume [tex]a\ge0[/tex] and for convenience that [tex]a<b[/tex]. Then
[tex]\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)[/tex]
