# Math Help - calc2. find the arc length of a function

1. ## calc2. find the arc length of a function

y=ln[(e^x +1)/(e^x -1)]
interval [ln2,ln3]

thanks

2. I assume you know the arc-lenght formula for this case.

3. y'= (-2e^x)/(e^2x - 1)
(y'^2)= [4e^(2x)]/[(e^2x - 1)^2]
sqroot of [1+ (y'^2)]= (e^2x +1) / (e^2x -1)

then i got stuck. i have no idea how to take the integral of (e^2x +1) / (e^2x -1)

thanks

4. Originally Posted by tichua

i have no idea how to take the integral of (e^2x +1) / (e^2x -1)
I'll assume your work is fine.

Now, to integrate that, just observe that $\frac{{{e}^{2x}}+1}{{{e}^{2x}}-1}=\frac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}$ and by differentiating $e^x-e^{-x}$ you'll get $e^x+e^{-x}$ which is the numerator, so I think you know what's the proper substitution.