# Arc length using integrals

• April 2nd 2009, 08:55 PM
bnay
Arc length using integrals
Ok, I've got this equation here,

y=ln((e^x+1)/(e^x-1)

and from this equation I need to find the arc length where p< x < q, p > 0.

I know the formula for the length is the integral from p to q of [1+(dy/dx)^2]^2dx, and I've tried u substitution eventually winding up with the integral from e^p to e^q [(u^2-1)/(u^2+1)*du/u, but here I'm stuck. Does anybody know where to go from here, or if I've just messed it up entirely?

Thanks
• April 2nd 2009, 10:16 PM
matheagle
I assume you want y=ln((e^x+1)/(e^x-1))

and

integral from p to q of [1+(dy/dx)^2]^{1/2}dx
• April 2nd 2009, 10:55 PM
bnay
Yep, its pretty much that integral that's stumped me
• April 2nd 2009, 11:01 PM
matheagle
You had left out the square root and the ).
BUT after some algebra I got

${dy\over dx}= {-2e^x\over e^{2x}-1}$, then the integrand becomes ${e^{2x}+1\over e^{2x}-1}$.

So you need to evaluate $\int_q^p {e^{2x}+1\over e^{2x}-1}dx=\int_q^p 1+{2\over e^{2x}-1} dx$.
• April 3rd 2009, 03:39 AM
Soroban
Hello, bnay!

I believe your final function is inverted . . .

Quote:

$y\:=\:\ln\left(\frac{e^x+1}{e^x-1}\right)$

I need to find the arc length for: $0 < p< x < q$

We have: . $y \;=\;\ln(e^x+1) - \ln(e^x-1)$

. . Then: . $\frac{dy}{dx} \;=\;\frac{e^x}{e^x+1} + \frac{e^x}{e^x-1} \;=\;\frac{-2e^x}{(e^{2x}-1)^2}$

And: . $1 + \left(\frac{dy}{dx}\right)^2 \;=\; 1 + \frac{4e^{2x}}{(e^{2x}-1)^2} \;=\;\frac{(e^{2x}-1)^2 + 4e^{2x}}{(e^{2x}-1)^2} \;=\;\frac{e^{4x} + 2x^{2x} + 1}{(e^{2x} - 1)^2}$ . $= \;\frac{(e^{2x} + 1)^2}{(e^{2x}-1)^2}$

. . Hence: . $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\frac{e^{2x}+1}{e^{2x}-1}$

And we must integrate: . $L \;=\;\int^p_q\frac{e^{2x}+1}{e^{2x}-1}\,dx$

Let: $u = e^x\quad\Rightarrow\quad x = \ln u \quad\Rightarrow\quad dx = \tfrac{du}{u}$

Substitute: . $L \;=\;\int \frac{u^2+1}{u^2-1}\,\frac{du}{u} \;=\;\int\frac{u^2+1}{u(u-1)(u+1)}\,du$

Partial Fractions: . $\int\left[-\frac{1}{u} + \frac{1}{u-1} + \frac{1}{u+1}\right]\,du$

I'll let you finish it . . .