What is the arc length of f(x)=(2x^2–ln(1/x+1)) on x in [1,2]?

Jun 24, 2018

$6.368995013$

Explanation:

Given
$f \left(x\right) = 2 {x}^{2} + \ln \left(\frac{1}{x} + 1\right)$
so we get

$f ' \left(x\right) = 4 x + \frac{1}{{x}^{2} \left(\frac{1}{x} + 1\right)}$
this is

$f ' \left(x\right) = 4 x + \frac{1}{x + {x}^{2}}$

$f ' \left(x\right) = \frac{4 x \left(x + {x}^{2}\right) + 1}{x + {x}^{2}}$

$f ' \left(x\right) = \frac{4 {x}^{3} + 4 {x}^{2} + 1}{x + {x}^{2}}$
and our integral will be

${\int}_{1}^{2} \sqrt{1 + {\left(\frac{4 {x}^{3} + 4 {x}^{2} + 1}{x \left(x + 1\right)}\right)}^{2}} \mathrm{dx}$
By a numerical method we get
$\approx 6.368995013$