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^2=2lnx # => #f'(x)=2/x#
Then
#L=int_1^3 sqrt(1+4/x^2)dx=int_1^3 sqrt(x^2+4)/xdx=F(x=3)-F(x=1)#
Making
#x=2tany#
#dx=2sec^2y*dy#
we get
#F(x)=int sqrt(x^2+4)/xdx=int(2secy*cancel2sec^2y)/(cancel2tany)dy=2int (1/cos^3y)(cosy/siny)dy=2int dy/(cos^2y*siny)#
But
#1/(cos^2y*siny)=(siny)/(cos^2y)+1/siny#
So
#F(x)=2int siny/cos^2ydy+2int dy/siny#
Making #cosy=z# => #siny*dy=-dz#
The first part becomes
#-2int dz/z^2=2/z=2/cosy#
Therefore
#F(x)=2/cosy+2ln |csc y-coty| +const.#
But
#x=2tany# => #siny=x/2cosy#
#sin^2y+cos^2y=1# => #(x^2/4+1)cos^2y=1# => #cosy=2/sqrt(x^2+4)#
#-> siny=x/2.(2/sqrt(x^2+4))# => #siny=x/sqrt(x^2+4)#
So
#F(x)=sqrt(x^2+4)+2ln |sqrt(x^2+4)/x-2/x|+const.#
#F(x)=sqrt(x^2+4)+2ln |sqrt(x^2+4)-2|-2ln|x|+const.#
Finally
#L=F(x=3)-F(x=1)=sqrt13+2ln (sqrt13-2)-2ln3-(sqrt5+2ln(sqrt5-2)-2ln1)#
#L=sqrt13-sqrt5+ln((sqrt13-2)/(3(sqrt5-2)))#
But
#(sqrt13-2)/(sqrt5-2)*(sqrt5+2)/(sqrt5+2)=sqrt65+2sqrt13-2sqrt5-4#
So
#L=sqrt13-sqrt5+ln((sqrt65+2sqrt13-2sqrt5-4)/3)~=3.006#
