How to find the arc length of this graph?

Could you please help me with this problem? Find the arc length of the graph of $y = \frac{x^{3}}{3} + \frac{1}{4x}$ between $x = 1$ and $x = 2$. Note: It may be helpful to use identities like $$x^{2} + \frac{1}{4x^{2}} = x^4 + \frac{1}{2} + \frac{1}{16x^4}.$$

The answer is $59/24$, but I have no idea how to obtain that. I get stuck after trying to use the identity and having to integrate $$\left( \frac{17}{18} + \frac{x^{4}}{81} + \frac{1}{16x^4}\right)^{1/2}.$$ Is it even possible to integrate this...?

Basically the arc length is given by : $$\int_a^b \sqrt{1+\left[f'(x)\right]^2} \ \mathrm{d}x$$ In your case $f'(x)=x^2-\frac{1}{4x^2}$, then \begin{aligned}\text{The arc length} &= \int_1^2 \sqrt{1+\left(x^2-\frac{1}{4x^2}\right)^2 } \ \mathrm{d}x\\ &= \int_1^2 \sqrt{\frac{(4x^4+1)^{2}}{16x^4} } \ \mathrm{d}x \\ &= \int_1^2x^2+\frac{1}{x^2} \ \mathrm{d}x \\ &= \frac{7}{3} +\frac{1}{8} = \frac{59}{24}. \end{aligned}
I think it is easy to see that while $y=x^3/3+1/4x$ then $$\sqrt{1+y'^2}=\frac{1}4\sqrt{\frac{(4x^4+1)^2}{x^4}}=\frac{4x^4+1}{4x^2}, ~ (1<x<2)$$