$f(x) = \frac{x^3}{6}+\frac{1}{2x}$, $\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$ $$f(x) =  \frac{x^3}{6}+\frac{1}{2x}$$
$$\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$$
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
$$\dfrac{x^4-3}{6x}+C = \boxed{\frac{14}{3}}$$

Note that
$$f(x) =  \frac{x^3}{6}+\frac{1}{2x}\implies f'(x)=\frac {x^2}{2}-\frac1{2x^2}$$
$$ \implies I=\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx =\int^{3}_{1} \sqrt{(\frac {x^2}{2}+\frac 1{2x^2})^2}\, dx $$
$$\implies I=\left . \frac {x^3}6-\frac 1 {2x}\right |^3_1 = \frac {14}3$$
And that you can't write this
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx = \dfrac{x^4-3}{6x}+C$$
since the integral is definite it's a number it can't be equal to a function...
