Question

Find the length of the arc of the curve $y=\log \left(\frac{e^x-1}{e^x+1}\right)$ from $x=1$ to $x=2$.

Answer 1

$y=\log \left(\frac{e^x-1}{e^x+1}\right)$ $x=1$ to $x=2$

$y={\int }_1^2\log \left(\frac{e^x-1}{e^x+1}\right)$

$y={\left[\frac{e^x+1}{e^x-1}\right]}_1^2$

$y=[\frac{e^2+1}{e^2-1}-\frac{e-1}{e-1}]$