#### Answer

$$\frac{13}{12}$$

#### Work Step by Step

Given
$$ y= y=\frac{1}{12} x^{3}+x^{-1}$$
Since \begin{aligned} 1+\left(y^{\prime}\right)^{2} &=1+\left(\frac{1}{4} x^{2}-x^{-2}\right)^{2} \\ &=1+\frac{1}{4} x^{4}-\frac{1}{2}+x^{-4} \\ &=\frac{1}{4} x^{4}+\frac{1}{2}+x^{-4} \\ &=\left(\frac{1}{4} x^{2}+x^{-2}\right)^{2} \end{aligned}
Then the arc length given by
\begin{aligned} s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\ &=\int_{1}^{2} \sqrt{\left(\frac{1}{4} x^{2}+x^{-2}\right)^{2}} d x \\ &=\int_{1}^{2}\left(\frac{1}{4} x^{2}+x^{-2}\right) d x \\ &=\left[\frac{1}{12} x^{3}-x^{-1}\right]_{1}^{2} \\ &=\frac{1}{6}+\frac{11}{12} \\ &=\frac{13}{12} \end{aligned}