Looking at this as a graph of a function of $y$ is more convenient $$ y=\frac{x^4+3}{6x}\Rightarrow \frac{dy}{dx}=\frac{x^3-1}{2x^2}\Rightarrow \left( \frac{dy}{dx} \right)^2=\frac{x^6-2x^3+1}{4x^4} $$ And the integral for arc length: $$ \int_1^2 \sqrt{1+\frac{x^6-2x^3+1}{4x^4}} \, \mathrm dx=\int_1^2 \frac{1}{2x^2} \sqrt{4x^4+x^6-2x^3+1} \, \mathrm dx $$ Which I am not sure how to evaluate. Substitutions don't seem fruitful and integration by parts looks like it will be messy. Did I set things up wrong? Is there an easy way to evaluate this integral?
edit: As Dr. MV pointed out, I made a differentiation error which he corrects below.
HINT:
$$1+y'^2(x)=\frac14\left(x^2+\frac{1}{x^2}\right)^2$$