Find arc length of implicit function Result of integrating I've been trying to find the arc length of the following function:
$e^y=\frac{e^x+1}{e^x-1}$
from $x = a$ to $x = b$. 
I've tried designating $y$, but it doesn't seem to work. I would really appreciate your help
 A: We have $y = $ ln$(\frac{e^x + 1}{e^x-1})$. Then $\frac{dy}{dx} = \frac{e^x -1}{e^x + 1} * \frac{{e^{2x} - e^x -e^{2x} - e^x}}{(e^x -1)^2} = -2\frac{e^x}{(e^x -1)(e^x +1)} = -2\frac{e^x}{e^{2x} -1}$.
Now, letting $y(x)$ be our curve and $l$ its length, $l(y) = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} dx$.
Then we have now $l(y) = \int_a^b \sqrt{1 + \frac{4e^{2x}}{e^{4x} - 2e^{2x} +1}}dx=\int_a^b \sqrt{\frac{e^{4x} +2e^{2x} + 1}{e^{4x} - 2e^{2x} +1}}dx=$ $ \int_a^b \sqrt{\frac{(e^{2x} +1)^2}{(e^{2x} -1)^2}}dx = \int_a^b \frac{e^{2x} +1}{e^{2x} -1}dx$.
Now put $u = e^{2x} -1$ Then $du = 2e^{2x}dx,$ or, $dx = \frac{du}{2(1+u)}$. For limits of integration, when $x = a, u = e^{2a} -1$, and when $x=b, u = e^{2b} -1$. 
Then our integral becomes $l(y) = \frac{1}{2}\int_{e^{2a} -1}^{e^{2b} -1}\frac{u+2}{u(1+u)}du = \frac{1}{2}\int_{e^{2a} -1}^{e^{2b} -1}\frac{1}{1+u} + \frac{2}{u(1+u)}du$.
Now, you can split up the integrand into two integrals. The first is quite simple, and the second is a routine application of partial fractions decomposition.
