What is the arclength of #f(x)=x/(x-5) in [0,3]#?

#f(x)=x/(x-5)#, #x in [0,3]#

Let #y=f(x)#. We can invert:

#x=5/(y-1)+5#, #y in [-3/2,0]#

Take the derivative with respect to #y#:

#x'=-5/(y-1)^2#

Arclength is given by:

#L=int_(-3/2)^0sqrt(1+25/(y-1)^4)dy#

Apply the substitution #1-y=u/2#:

#L=1/2int_2^5sqrt(1+400/u^4)du#

Complete the square:

#L=1/2int_2^5sqrt((1+20/u^2)^2-40/u^2)du#

Factor out the larger piece:

#L=1/2int_2^5(1+20/u^2)sqrt(1-(40u^2)/(u^2+20)^2)du#

For #u in [2,5]#, #(40u^2)/(u^2+20)^2<1#. Take the series expansion of the square root:

#L=1/2int_2^5(1+20/u^2){sum_(n=0)^oo((1/2),(n))(-(40u^2)/(u^2+20)^2)^n}du#

Isolate the #n=0# term and simplify:

#L=1/2int_2^5(1+20/u^2)du+1/2sum_(n=1)^oo((1/2),(n))(-40)^nint_2^5 1/u(u/(u^2+20))^(2n-1)du#

Apply the substitution #u=sqrt20tantheta#:

#L=1/2[u-20/u]_ 2^5+sqrt5sum_(n=1)^oo((1/2),(n))(-2)^nint((tantheta)/(sec^2theta))^(2n-2)d theta#

Isolate the #n=1# term and simplify:

#L=9/2-sqrt5intd theta+sqrt5sum_(n=2)^oo((1/2),(n))(-2)^nint((tantheta)/(sec^2theta))^(2n-2)d theta#

Apply the appropriate half-angle Trigonometric identities:

#L=9/2-sqrt5[tan^(-1)(u/sqrt20)]_ 2^5+4sqrt5sum_(n=2)^oo((1/2),(n))((-1)/2)^nintsin^(2n-2)(2theta)d theta#

Rescale #n#:

#L=9/2-sqrt5(tan^(-1)(sqrt5/2)-tan^(-1)(1/sqrt5))-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/2)^nintsin^(2n)(2theta)d theta#

Apply the appropriate Trigonometric power-reduction formula:

#L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/2)^nint{1/4^n((2n),(n))+2/4^nsum_(m=0)^(n-1)(-1)^(n-m)((2n),(m))cos((4n-4m)theta)}d theta#

Integrate term by term:

#L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/8)^n[((2n),(n))theta+1/2sum_(m=0)^(n-1)(-1)^(n-m)/(n-m)((2n),(m))sin((4n-4m)theta)]#

Reverse the last substitution:

#L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/8)^n[((2n),(n))tan^(-1)(u/sqrt20)+1/2sum_(m=0)^(n-1)(-1)^(n-m)/(n-m)((2n),(m))sin((4n-4m)tan^(-1)(u/sqrt20))]_2^5#

Insert the limits of integration:

#L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/8)^n{((2n),(n))cot^(-1)(sqrt5)+1/2sum_(m=0)^(n-1)(-1)^(n-m)/(n-m)((2n),(m))(sin((4n-4m)tan^(-1)(sqrt5/2))-sin((4n-4m)tan^(-1)(1/sqrt5)))}#

Apply the appropriate Trigonometric angle-addition identities:

#L=9/2-sqrt5cot^(-1)(sqrt5)-2sqrt5sum_(n=1)^oo((1/2),(n+1))(-1/8)^n{((2n),(n))cot^(-1)(sqrt5)+sum_(m=0)^(n-1)(-1)^(n-m)/(n-m)((2n),(m))sin((2n-2m)cot^(-1)(sqrt5))cos((2n-2m)tan^(-1)(4/sqrt5))}#