Find the arc length of the curve y = ln (e^x - 1)/(e^x +

Find the arc length of the curve $y = \ln\dfrac{e^x - 1}{e^x + 1}$ from the point $M_1(a, y_1)$ to the point $M_2(b, y_2)$, where $0 \lt a \lt b$.

The curve requires $e^x>1$ (i.e. $x>0$), which holds throughout $[a,b]$ since $0 \lt a \lt b$.

Step 1: Differentiate

\[y=\ln(e^x-1)-\ln(e^x+1) \implies y'=\frac{e^x}{e^x-1}-\frac{e^x}{e^x+1}=\frac{2e^x}{e^{2x}-1}\]

Step 2: Simplify $1+(y')^2$

\[1+(y')^2=\frac{(e^{2x}-1)^2+4e^{2x}}{(e^{2x}-1)^2}=\frac{e^{4x}+2e^{2x}+1}{(e^{2x}-1)^2}=\frac{(e^{2x}+1)^2}{(e^{2x}-1)^2}\]

Since $x>0$ means $e^{2x}>1$, both terms are positive, so

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

Step 3: Decompose the integrand

\[\frac{e^{2x}+1}{e^{2x}-1}=\frac{(e^{2x}-1)+2}{e^{2x}-1}=1+\frac{2}{e^{2x}-1}\]

Step 4: Integrate the remaining term

With $u=e^{2x}-1$ (so $du=2e^{2x}dx=2(u+1)dx$) and partial fractions $\dfrac{1}{u(u+1)}=\dfrac1u-\dfrac1{u+1}$:

\[\int\frac{2}{e^{2x}-1}\,dx=\int\frac{du}{u(u+1)}=\ln|u|-\ln|u+1|=\ln(e^{2x}-1)-2x\]

Step 5: Combine the antiderivative

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

Step 6: Evaluate at the limits

\[L=\big[\ln(e^{2x}-1)-x\big]_a^b=\big[\ln(e^{2b}-1)-b\big]-\big[\ln(e^{2a}-1)-a\big]\]

\[=\ln\frac{e^{2b}-1}{e^{2a}-1}-(b-a)\]

\[\boxed{L=\ln\frac{e^{2b}-1}{e^{2a}-1}-(b-a)}\]

Step 7: Sanity check

Since $b>a>0$, $e^{2b}-1>e^{2a}-1$, so the logarithm is positive; a numerical check at $a=1,\,b=2$ gives $L\approx1.127$, matching direct numerical integration of $\int_1^2\sqrt{1+(y')^2}\,dx$.