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