Question

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

Answer 1

`1.000153728`

Explanation:

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`