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

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Answer 1 · 2018-05-16T06:49:10

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

#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))#