`y = ln((e^x + 1)/(e^x - 1)) , [ln2 , ln3]` Find the arc length of the graph of the function over the indicated interval.

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The arc length of a function is the length of the described curve within some interval on the x-axis (and a corresponding interval on the y-axis). If we were to lay a piece of string over the line of the graph in this window, the graph going from corner to...

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The arc length of a function is the length of the described curve within some interval on the x-axis (and a corresponding interval on the y-axis). If we were to lay a piece of string over the line of the graph in this window, the graph going from corner to diagonally opposite corner in a curve, the arc length is the length of the piece of string used.

In this case the function to find the arc length of is

`y = ln ((e^x+1)/(e^x - 1)) `

where `ln(x) ` is the natural logarithm of `x ` (the inverse function to `e^(x) `, recall).

The window over which to find the length is the rectangle given by  `ln 2 <= x <= ln3 `  and `ln2 <= y <= ln3`.

The function for the arc length of a generic function `y = f(x) ` is

`s = int_a^b sqrt(1+((dy)/(dx))^2) \quad dx `

This integrates along the line of the graph, adding infinitesimal sections together where each very small section added is a straight line diagonal over the small window `x_0 <=x <=x_0 + dx ` , `y_0 <=y <= y_0 + dy ` . Though `y = f(x) ` may be a curve, the point is that these windows over which the length is integrated (added up) are so small that the graph is a straight line within them making it a simple thing to add the small (straight line) sections together. This concept is the basis of calculus.

In this example, since `y = ln ((e^x+1)/(e^x-1)) = ln(e^x + 1) - ln(e^x-1)` then its derivative `(dy)/(dx) `  is given by

`(dy)/(dx) = (1/(e^x+1))e^x - (1/(e^x-1))e^x = e^x(1/(e^x+1) - 1/(e^x-1)) ` `= e^x((e^x-1-e^x-1)/((e^x+1)(e^x-1))) ` `= -(2e^x)/((e^x+1)(e^x-1))`

Now working in steps to find the integrand in the formula for the arc length `s ` , we have that

`1+ ((dy)/(dx))^2 = 1 + (4e^(2x))/((e^x+1)^2(e^x-1)^2) `  `= ((e^x+1)^2(e^x-1)^2 + 4e^(2x))/((e^x+1)^2(e^x-1)^2) `

and that

`sqrt(1+((dy)/(dx))^2) = (sqrt((e^x+1)^2(e^x-1)^2 + 4e^(2x)))/((e^x+1)(e^x-1)) `

`= sqrt((e^(2x)+2e^x +1)(e^(2x)-2e^(x)+1)+4e^(2x))/((e^x+1)(e^x-1)) `

`= sqrt((e^(4x)-2e^(3x)+e^(2x)+2e^(3x) - 4e^(2x) + 2e^x + e^(2x)-2e^x + 1)+4e^(2x))/((e^x+1)(e^x-1)) `

Adding up that very long set of exponential terms, finding that most cancel, we finally have that

`sqrt(1+((dy)/(dx))^2) = sqrt((e^(4x)-2e^(2x)+1)+4e^(2x))/((e^x+1)(e^x-1)) `  `= sqrt(e^(4x)+2e^(2x)+1)/((e^x+1)(e^x-1)) `

`= sqrt((e^(2x)+1)^2)/((e^x+1)(e^x-1)) = (e^(2x)+1)/((e^x+1)(e^x-1)) `

`= (e^(2x)+1)/(e^(2x)-1) `

Integrating the integrand over the required interval, we have that

`s = int_(ln2)^(ln3) ((e^(2x)+1)/(e^(2x)-1)) \quad dx `

` ` `= int_(ln2)^(ln3) (2e^(2x)- e^(2x) + 1)/(e^(2x)-1) \quad dx = int_(ln2)^(ln3) (2e^(2x))/(e^(2x)-1) -1 \quad dx `

`= ln(e^(2x)-1)|_ln2^(ln3) - x |_(ln2)^(ln3) `

`= ln (e^(2ln3)-1) - ln(e^(2ln2)-1) - ln3 + ln2 `

`= ln(9-1) - ln(4-1) - ln3 + ln2 = ln((8/3)(2/3)) `

`= ln(16/9)`  is the final answer

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