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

Question

$$f(x) = \frac{x^3}{6}+\frac{1}{2x}$$

$$\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$$

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?

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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

$$\dfrac{x^4-3}{6x}+C = \boxed{\frac{14}{3}}$$

Answer 1

Hint: Note that $$1+\left(\frac{x^4-1}{2x^2}\right)^2=\frac{(x^4+1)^2}{4x^4}$$

Answer 2

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...