$$y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$$
$$\frac{dy}{dx}=\frac{5}{6}x^{4}-\frac{3}{10x^{4}}$$
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.