What is the arc length of f(x)= lnx on x in [1,3] ?

1 Answer
Mar 16, 2016

L=sqrt10-sqrt(2)+ln((2sqrt5+sqrt10-sqrt2-1)/3)~=2.302

Explanation:

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)