Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 61

Arc length of curve can be denoted as "S ". We can determine it by using integral formula on a closed interval [a,b] as: S = int_a^b ds
where:
ds = sqrt(1+ ((dy)/(dx))^2 )dx if y=f(x)
or
ds = sqrt(1+((dx)/(dy))^2) dy if x=h(y)
a = lower boundary of the closed interval
b =upper boundary of the closed interval

From the given problem: y =ln(x), [1,5] , we determine that the boundary values are:
a= 1 and b=5
Note that y= ln(x) follows y=f(x) then the formula we will follow can be expressed as S =int_a^bsqrt(1+ ((dy)/(dx))^2 )dx
For the derivative of y or (dy)/(dx) , we apply the derivative formula for logarithm:
d/(dx)y= d/(dx) ln(x)
(dy)/(dx)= 1/x
Then ((dy)/(dx))^2= (1/x)^2 or 1/x^2 .
Plug-in the values on integral formula for arc length of a curve, we get:
S =int_1^5sqrt(1+1/x^2 )dx
Let 1 = x^2/x^2 then we get:
S=int_1^5sqrt(x^2/x+1/x^2 )dx
=int_1^5sqrt((x^2+1)/x^2 )dx
=int_1^5sqrt(x^2+1)/sqrt(x^2 )dx
=int_1^5sqrt(x^2+1)/sqrt(x^2 )dx
=int_1^5sqrt(x^2+1)/xdx
From the integration table, we follow the formula for rational function with roots:
int sqrt(x^2+a^2)/x dx = sqrt(x^2+a^2)-a*ln|(a+sqrt(x^2+a^2))/x| .
Applying the integral formula with a^2=1 then a=1, we get:
int_1^5sqrt(x^2+1)/xdx = [sqrt(x^2+1)-1*ln|(1+sqrt(x^2+1))/x|]|_1^5
= [sqrt(x^2+1)-ln|(1+sqrt(x^2+1))/x|]|_1^5
Apply the definite integral formula: F(x)|_a^b= F(b)-F(a) .
[sqrt(x^2+1)-ln|(1+sqrt(x^2+1))/x|]|_1^5
=[sqrt(5^2+1)-ln|(1+sqrt(5^2+1))/5|]-[sqrt(1^2+1)-ln|(1+sqrt(1^2+1))/1|]
=[sqrt(25+1)-ln|(1+sqrt(25+1))/5|]-[sqrt(1+1)-ln|(1+sqrt(1+1))/1|]
=[sqrt(26)-ln|(1+sqrt(26))/5|]-[sqrt(2)-ln|1+sqrt(2)|]
=sqrt(26)-ln|(1+sqrt(26))/5| -sqrt(2)+ln|1+sqrt(2)|
Apply logarithm property: ln(x)-ln(y) = ln(x/y) .
S =sqrt(26)-sqrt(2)+ln|1+sqrt(2)|-ln|(1+sqrt(26))/5|
S =sqrt(26)-sqrt(2)+ln|(1+sqrt(2))/(((1+sqrt(26))/5))|
S =sqrt(26)-sqrt(2)+ln|(5*(1+sqrt(2)))/(1+sqrt(26))|
S =sqrt(26)-sqrt(2)+ln|(5+5sqrt(2))/(1+sqrt(26))|
S~~4.37