Question

Find the length of the curve `y=ln(e^(x)-1/e^(x)+1)` from `x=1` to `x=2`?

Answer 1

`L ~~ 0.432333...`

Explanation:

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

Rewrite as:

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

Use the chain rule, `(dy(g))/dx = dy/(dg)(dg)/dx`, to compute the derivative:

We observe that `y = ln(g) and g = e^x-e^-x+1`, then `dy/(dg) = 1/g = 1/(e^x-e^-x+1)` and `(dg)/dx = e^x+e^-x`

Substituting into the chain rule:

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

Multiply the right side by 1 in the form of `e^x/e^x`:

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

The curve length is:

`L = int_a^b sqrt(1- (dy/dx)^2)dx`

Substitute the derivative and the limits:

`L = int_1^2 sqrt(1- ((e^(2x)+1)/(e^(2x)+e^x-1))^2)dx`

The indefinite integral cannot be resolved using standard functions, therefore, I used WolframAlpha to perform a numeric integration:

`L = int_1^2 sqrt(1- ((e^(2x)+1)/(e^(2x)+e^x-1))^2)dx ~~ 0.432333...`