Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ $$y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$$
$$\frac{dy}{dx}=\frac{5}{6}x^{4}-\frac{3}{10x^{4}}$$
squaring this $$=\frac{25}{36}x^{8}+\frac{9}{100x^{8}}$$
Plugging into the formula $$ds=\sqrt{1+\left( \frac{dy}{dx}\right) ^{2}}$$
$$\int_{1}^{2}\sqrt{1+\frac{25}{36}x^{8}+\frac{9}{100x^{8}}}$$
Is this correct so far? And how would I go about evaluating this integral.

As Américo Tavares pointed out, you are missing the term $-1/2$ due to the crossterm when squaring $d y/ dx$. The integral you want to solve is
$$\int_1^2 dx \,\sqrt{\frac{1}{2}+ \frac{25 x^8}{36} + \frac{9}{100 x^8}}
= \int_1^2 dx \, \sqrt{\frac{(9+25 x^8)^2}{900 x^8}}$$
$$ = \int_1^2 dx \,\frac{9+25 x^8}{30 x^4} = \left[-\frac{1}{10 x^3} + \frac{x^4}{6} \right]_{x=1}^2 =  \frac{1261}{240}. $$
I hope every step is reproducible.
