To do this we need to apply the formula for the length of the curve
mentioned in:
How do you find the length of a curve using integration?
We start from
#L=int_a^b sqrt(1+(f' (x))^2 )dx#
#f(x)=lnx# => #f'(x)=1/x#
Then
#L=int_1^3 sqrt(1+1/x^2)dx=int_1^3 sqrt(x^2+1)/xdx#
Making
#x=tany#
#dx=sec^2y*dy#
we get
#int sqrt(x^2+1)/xdx=int(secy*sec^2y)/tanydy=int (1/cos^3y)(cosy/siny)dy=int dy/(cos^2y*siny)#
We can break the integrand in partial fractions in this way
#1/((1-sin^2x)*siny)=1/((1-siny)(1+siny)siny)=A/(1+siny)+B/(1-siny)+C/siny# [1]
For #y=pi/6 and (5pi)/6#, we get
#8/3=2/3A+2B+2C#
#-8/3=2A+2/3B-2C# [2]
Summing the last two expressions we get
#0=8/3A+8/3B# => #B=-A#
Then expression [2] becomes
#-8/3=4/3A-2C# => #-4/3=2/3A-C# [3]
Since (in expression [1]) #A/(1+siny)-A/(1-siny)=A*(1-siny-1-siny)/cos^2y=-(2Asiny)/cos^2y#
the integrand can be rewritten as
#1/(cos^2y*siny)=-(2Asiny)/cos^2y+C/siny#
For #y=pi/4#
#1/((sqrt2/2)^2*cancel(sqrt2/2))=-(2Asqrt2/2)/(sqrt2/2)^cancel(2)+C/(cancel(sqrt2/2))#
#2=-2A+C#
Summing the last expression with expression [3] we get
#2/3=-4/3A# => #A=-1/2#
#-> C=2+2A=2-1# => #C=1#
So we arrived at
#=int siny/cos^2ydy+int dy/siny#
Making #cosy=z# => #siny*dy=-dz#
The first part becomes
#int -dz/z^2=1/z=1/cosy#
Therefore
#=1/cosy+ln |csc y-coty| +const.#
But
#x=tany# => #siny=xcosy#
#sin^2y+cos^2y=1# => #(x^2+1)cos^2y=1# => #cosy=1/sqrt(x^2+1)#
#-> siny=x.(1/sqrt(x^2+1))# => #siny=x/sqrt(x^2+1)#
So
#=sqrt(x^2+1)+ln |sqrt(x^2+1)/x-1/x|+const.#
#=sqrt(x^2+1)+ln |(sqrt(x^2+1)-1)/x|+const.#
Finally
#L=(sqrt(x^2+1)+ln |(sqrt(x^2+1)-1)/x|)|_1^3#
#L=sqrt10+ln ((sqrt10-1)/3)-sqrt2-ln(sqrt2-1)#
#L=sqrt10-sqrt2+ln((sqrt10-1)/(3(sqrt2-1)))#
But
#(sqrt10-1)/(sqrt2-1)*(sqrt2+1)/(sqrt2+1)=2sqrt5+sqrt10-sqrt2-1#
So
#L=sqrt10-sqrt2+ln((2sqrt5+sqrt10-sqrt2-1)/3)#
