Differentiate r(x) :
[tex]r'(x) = \left(12x,2\sqrt 6,\dfrac1x\right)[/tex]
Then the arc length of r(x) over 1 ≤ x ≤ 6 is given by the integral,
[tex]\displaystyle \int_1^6 \|r'(x)\|\,\mathrm dx = \int_1^6 \sqrt{(12x)^2 + (2\sqrt6)^2 + \left(\frac1x\right)^2} \,\mathrm dx = \int_1^6 \sqrt{144x^2 + 24 + \frac1{x^2}}\,\mathrm dx[/tex]
Notice that the expression in the square root is a perfect square:
[tex]144x^2 + 24 + \dfrac1{x^2} = (12x)^2 + 2\cdot12x\cdot\dfrac1x + \left(\dfrac1x\right)^2 = \left(12x+\dfrac1x\right)^2[/tex]
Then in the integral, we have
[tex]\displaystyle \int_1^6 \sqrt{\left(12x+\frac1x\right)^2}\,\mathrm dx = \int_1^6 \left|12x + \frac1x\right|\,\mathrm dx[/tex]
but the integrand is positive over its entire domain, so we can drop the absolute value.
The arc length is then
[tex]\displaystyle \int_1^6 \left(12x+\frac1x\right)\,\mathrm dx = (6x^2 + \ln|x|)\bigg|_1^6 = (6^3+\ln(6)) - (6+\ln(1)) = \boxed{210+\ln(6)}[/tex]
