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

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Answer 1 · 2018-06-09T12:12:16

$1.000153728$

Wrting your $f(x)$ in the form

$f(x)=1/sqrt(2)*1/sqrt(x^2-1)$

then $f'(x)$ is given by

$f'(x)=-1/2*sqrt(2)x/(x^2-1)^(3/2)$
then our integralo is given by

$1/2*int_6^7sqrt(4+2*x^2/(x^2-1)^3)dx$
A numerical method gives
$approx 1.000153728$

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