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

1 Answer
May 16, 2018

L=1+sum_(n=1)^oo((1/2),(n))1/(4n-1)(1/2^(4n-1)-1/3^(4n-1))

Explanation:

f(x)=(x-2)/(x^2-x-2)=1/(x+1)

f'(x)=-1/(x+1)^2

Arclength is given by:

L=int_1^2sqrt(1+1/(x+1)^4)dx

For x in [1,2], 1/(x+1)^4<1. Take the series expansion of the square root:

L=int_1^2sum_(n=0)^oo((1/2),(n))1/(x+1)^(4n)dx

Isolate the n=0 term and simplify:

L=int_1^2dx+sum_(n=1)^oo((1/2),(n))int_1^2 1/(x+1)^(4n)dx

Integrate directly:

L=1+sum_(n=1)^oo((1/2),(n))1/(1-4n)[1/(x+1)^(4n-1)]_1^2

Insert the limits of integration:

L=1+sum_(n=1)^oo((1/2),(n))1/(4n-1)(1/2^(4n-1)-1/3^(4n-1))