Calculus (3rd Edition)

$$\ln \sinh (3)-\ln \sinh (1)$$
Since $$y=\ln \frac{e^x+1}{e^x-1}\to f(x)=\frac{e^x+1}{e^x-1}$$ Then \begin{aligned} f(x)^{2}+f^{\prime}(x)^{2} &=\frac{\left(e^{x}+1\right)^{2}}{\left(e^{x}-1\right)^{2}}+\left(-\frac{2 e^{x}}{\left(e^{x}-1\right)^{2}}\right)^{2} \\ &=\frac{\left(e^{x}+1\right)^{2}}{\left(e^{x}-1\right)^{2}}+\frac{4 e^{2 x}}{\left(e^{x}-1\right)^{4}} \\ &=\frac{\left(e^{x}-1\right)^{4}\left(e^{x}+1\right)^{2}+4 e^{2 x}\left(e^{x}-1\right)^{2}}{\left(e^{x}-1\right)^{2}\left(e^{x}-1\right)^{4}} \\ &=\frac{\left(e^{x}-1\right)^{2}\left(\left(e^{x}+1\right)^{2}\left(e^{x}-1\right)^{2}+4 e^{2 x}\right)}{\left(e^{x}-1\right)^{6}} \\ &=\frac{\left(e^{x}+1\right)^{2}\left(e^{x}-1\right)^{2}+4 e^{2 x}}{\left(e^{x}-1\right)^{4}}\\ &=\frac{(e^2x+1)^2}{(e^x-1)^4} \end{aligned} and \begin{aligned} s=\int_{a}^{b} \frac{\sqrt{f(x)^{2}+f^{\prime}(x)^{2}}}{f(x)} d x &=\int_{1}^{3} \frac{\sqrt{\frac{\left(e^{2 x}+1\right)^{2}}{\left(e^{x}+1\right)^{2}}}}{\left(\frac{\left.e^{x}-1\right)^{4}}{\left(e^{x}-1\right)^{4}}\right.} d x \\ &=\int_{1}^{3} \frac{e^{2 x}+1}{\left(e^{x}-1\right)} d x \\ &=\int_{1}^{3} \frac{\left(e^{2 x}+1\right)\left(e^{x}-1\right)}{\left(e^{x}-1\right)^{2}} \cdot \frac{e^{x}-1}{e^{x}+1} d x \\ &=\int_{1}^{3} \frac{e^{2 x}+1}{\left(e^{x}-1\right)^{2}\left(e^{x}+1\right)} d x \\ &=\int_{1}^{3} \frac{e^{2 x}+1}{e^{2 x}-1} d x \\ &=\int_{1}^{3} \frac{e^{2 x}+1}{e^{2 x}-1} d x\\ &=\int_{1}^{3} \frac{\frac{e^x+e^{-x}}{2}}{\frac{e^x-e^{-x}}{2}} d x\\ &=\int_{1}^{3}\coth xdx\\ &=\ln \sinh x\bigg|_{1}^{3}\\ &=\ln \sinh (3)-\ln \sinh (1) \end{aligned}