Can anyone find the arc length of the curve y=ln((e^x+1)/(e^x-1))?

Question

i need arc length using integration by parts.

it is (e^x)+1 and the same for the bottom

Answer 1

y = Ln((e^x +1)/(e^x - 1))
y =Ln(e^x + 1) - Ln(e^x-1)
ds/dx = √(1 + dy/dx)
dy/dx = e^x/(e^x + 1) - e^x/(e^x - 1)
dy/dx = [ e^x (e^x - 1) - e^x (e^x + 1) ] / (e^x + 1)(e^x - 1)
dy/dx = [ e^2x - 1e^x - e^2x + 1e^x ] / ( (e^x + 1) (e^x(e^x + 1) - 1(e^x + 1)) )
dy/dx = ( e^2x - e^2x ) / ( (e^2x + 1e^x) - 1(e^2x + 1e^x) )
dy/dx = ( e^2x - e^2x ) / ( e^2x - e^2x )
dy/dx = 1
∫ds = ∫(√2)dx
s = (√2)x + C
Let s(x0) = 0
then C = - (√2)x0
s = (√2)x - (√2)x0
s = (√2)(x - x0)

Answer 2

The formula for the arc length of a curve is as follows:

∫[a,b] √1+(dy/dx)^2 dx, where we are integrating between [a,b]. Since you did not provide a boundary, we can not give you a complete answer but you may use the F.T.C. to fill in the blanks.

The first step is to find dy/dx as it is a necessary component:
f(x) = ln((e^x+1)/(e^x-1)) so f'(x) = 0
**assuming you meant e^(x+1).., if not it will complicate things**

***EDIT***
Ok so you didn't mean e^(x+1).. thats ok it will just be a bit messier

We still need to find dy/dx of y= ln((e^x+1)/(e^x-1)) so I will just give it to you, but you should know how to do it on your own:

dy/dx = -1/(e^+1) - 1/(e^x-1) **this should look slightly familiar**

now the integral:

∫√1+(dy/dx)^2 dx .. can be simplified to:

∫ [a,b](e^(2x)+1)* |1/(e^(2x)-1)| dx

and then the integral gets crazy!! Good luck with this one!!

Answer 3

First we would desire to correctly known that the formula for arc length is, A=indispensable of sqrt(a million+(dy/dx)^2). to discover dy/dx, we take the derivative of y=2x-3. that's dy/dx=2. Than plug in 2 for dy/dx. which you get A=indispensable of sqrt(a million+(2)^2) => A=indispensable of sqrt(5) The indispensable of sqrt (5)= sqrt(5)*x than with the FTC(user-friendly theorem of calc) evaluate sqrt(5)*x from -3 to a million. As for the the checking area, it is going to be hardship-free, im beneficial you comprehend the dist formula.