<s>[INST] $f(x) = \frac{x^3}{6}+\frac{1}{2x}$, $\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$ $$f(x) =  \frac{x^3}{6}+\frac{1}{2x}$$
$$\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$$
Let's start by deriving the function, we have
$$f'(x) = \dfrac{x^4-1}{2x^2}$$
Hence we get
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx = ?$$
Am I right? 

UPDATE: 
If we have a definite integral: 
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx = \dfrac{x^4-3}{6x}+C$$
Then
$$\dfrac{x^4-3}{6x}+C = \boxed{\frac{14}{3}}$$
 [/INST] Hint: Note that $$1+\left(\frac{x^4-1}{2x^2}\right)^2=\frac{(x^4+1)^2}{4x^4}$$
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