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`

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