# Thread: another arc length question

1. ## another arc length question

Find the length of the curve, L.

not exactly sure how to integrate this one after the derivative is found.. which is very long and complicated..

2. Originally Posted by khood
Find the length of the curve, L.

not exactly sure how to integrate this one after the derivative is found.. which is very long and complicated..
$y=\ln \left( \frac{e^x+1}{e^x-1} \right)=\ln(e^x+1)-\ln(e^x-1)$

$y'=\frac{e^x}{e^x+1}-\frac{e^x}{e^x-1}=\frac{e^{x}(e^x-1)-e^{x}(e^x+1)}{(e^{x}+1)(e^{x}-1)}=\frac{-2e^x}{(e^x+1)(e^x-1)}$

$
\int \sqrt{1+\left( \frac{-2e^x}{(e^x+1)(e^x-1)}\right)^2}dx=\int \sqrt{\frac{[(e^x+1)(e^x-1)]^2+4e^{2x}}{[(e^x+1)(e^x-1)]^2}}dx
$

$\int \sqrt{\frac{[e^{2x}-1]^2+4e^{2x}}{[(e^x+1)(e^x-1)]^2}}dx=\int \sqrt{\frac{[e^{4x}-2e^{2x}+1+4e^{2x}}{[(e^x+1)(e^x-1)]^2}}dx$

$\int \sqrt{\frac{e^{4x}+2e^{2x}+1}{[(e^x+1)(e^x-1)]^2}}dx = \int \sqrt{ \frac{[e^{2x}+1]^2}{[(e^x+1)(e^x-1)]^2}}dx$

$\int \frac{e^{2x}+1}{(e^{x}+1)(e^{x}-1)}dx =\int \frac{e^{2x}+1}{e^{2x}-1}dx$

from here let $u = e^{2x} \implies du=2e^{2x}dx \iff du=2udx \iff \frac{du}{2u}=dx$

$\int \frac{u+1}{u-1}\frac{du}{2u}=\int\frac{u+1}{u(u-1)}du$ this can be integrated using partial fractions. Good luck.