What is the arclength of f(x)=(x^2+24x+1)/x^2  in the interval [1,3]?

Jul 12, 2018

$\approx 17.060431259$

Explanation:

Given $f \left(x\right) = \frac{{x}^{2} + 24 x + 1}{x} ^ 2$
we differentiate this function with respect to $x$ using the Quotient rule

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

so we get

$f ' \left(x\right) = \frac{\left(2 x + 24\right) {x}^{2} - \left({x}^{2} + 24 x + 1\right) 2 x}{x} ^ 4$
cancelling $x \ne 0$

$f ' \left(x\right) = \frac{2 {x}^{2} + 24 x - 2 {x}^{2} - 48 x - 2}{x} ^ 3$

$f ' \left(x\right) = \frac{- 2 \left(12 x + 1\right)}{x} ^ 3$

so we get

${\int}_{1}^{3} \sqrt{1 + {\left(- 2 \frac{12 x + 1}{x} ^ 3\right)}^{2}} \mathrm{dx}$
by a numerical method we get $17.06031259$