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

1 Answer

Jun 3, 2018

#approx 1.07667#

Explanation:

We get by the Quotient rule

#f'(x)=(2x(x-1)-(1+x^2))/(x-1)^2#
#f'(x)=(x^2-2x-1)/(x-1)^2#
so our integral is given by

#int_2^3sqrt(1+((x^2-2x-1)/(x-1)^2)^2)dx#

A numerical result is given by #approx 1.07667#

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