[tex]\mathbf r(t)=\langle6t^2,2\sqrt6 t,\ln t\rangle[/tex]
[tex]\implies\dfrac{\mathrm d\mathbf r}{\mathrm dt}=\left\langle12t,2\sqrt6,\dfrac1t\right\rangle[/tex]
The arc length is then given by
[tex]\displaystyle\int_1^4\sqrt{\frac{\mathrm d\mathbf r}{\mathrm dt}\cdot\frac{\mathrm d\mathbf r}{\mathrm dt}}\,\mathrm dt[/tex]
[tex]\displaystyle\int_1^4\sqrt{144t^2+24+\frac1{t^2}}\,\mathrm dt[/tex]
[tex]\displaystyle\int_1^4\sqrt{\left(12t+\frac1t\right)^2}\,\mathrm dt[/tex]
[tex]\displaystyle\int_1^4\left(12t+\frac1t\right)\,\mathrm dt[/tex]
[tex]=90+\ln4[/tex]
