#### Answer

$$\frac{779}{240}$$

#### Work Step by Step

Given $$y=\frac{x^{5}}{10}+\frac{x^{-3}}{6},\ \ \ [1,2]$$
Since
\begin{aligned}
y'&= \frac{x^{4}}{2}-\frac{1}{2 x^{4}}\\
\end{aligned}
Then the arc length is given by
\begin{aligned}
s&=\int_{a}^{b} \sqrt{1+y'^{2}} d x \\
&=\int_{1}^{2} \sqrt{1+\frac{x^{8}}{4}-\frac{1}{2}+\frac{1}{4 x^{8}} } d x \\
&=\int_{1}^{2} \sqrt{\frac{x^{8}}{4}+\frac{1}{2}+\frac{1}{4 x^{8}}} d x \\
&=\int_{1}^{2} \sqrt{\left(\frac{x^{4}}{2}+\frac{1}{2 x^{4}}\right)^{2}} d x \\
&=\int_{1}^{2} \sqrt{\left(\frac{x^{4}}{2}+\frac{1}{2 x^{4}}\right)^{2}} d x \\
&=\int_{1}^{2}\left(\frac{x^{4}}{2}+\frac{1}{2 x^{4}}\right) d x \\
&=\frac{1}{2} \int_{1}^{2}\left(x^{4}+x^{-4}\right) d x \\
&=\frac{1}{2}\left[\frac{x^{5}}{5}-\frac{x^{-3}}{3}\right]_{1}^{2}\\
&= \frac{779}{240}
\end{aligned}