# What is the arclength of f(x)=1/sqrt((x-1)(2x+2)) on x in [6,7]?

Jun 9, 2018

$1.000153728$

#### Explanation:

Wrting your $f \left(x\right)$ in the form

$f \left(x\right) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{{x}^{2} - 1}}$

then $f ' \left(x\right)$ is given by

$f ' \left(x\right) = - \frac{1}{2} \cdot \sqrt{2} \frac{x}{{x}^{2} - 1} ^ \left(\frac{3}{2}\right)$
then our integralo is given by

$\frac{1}{2} \cdot {\int}_{6}^{7} \sqrt{4 + 2 \cdot {x}^{2} / {\left({x}^{2} - 1\right)}^{3}} \mathrm{dx}$
A numerical method gives
$\approx 1.000153728$