Answer
$\ln(\frac{16}{9})$
Work Step by Step
$\frac{dy}{dx}$ = $\frac{e^{x}}{e^{x}-1}-\frac{e^{x}}{e^{x}+1}$
$\frac{dy}{dx}$ = $\frac{2e^{x}}{e^{2x}-1}$
$L$ = $\int_{{\,\ln2}}^{{\,\ln3}}$$\sqrt {1+(\frac{2e^{x}}{e^{2x}-1})^{2}}$ $dx$
$L$ = $\int_{{\,\ln2}}^{{\,\ln3}}$$\sqrt {\frac{e^{4x}+2e^{2x}+1}{(e^{2x}-1)^{2}}}$ $dx$
$L$ = $\int_{{\,\ln2}}^{{\,\ln3}}$$\sqrt {(\frac{e^{2x}+1}{e^{2x}-1})^{2}}$ $dx$
$L$ = $\int_{{\,\ln2}}^{{\,\ln3}}$${\frac{e^{2x}+1}{e^{2x}-1}}$ $dx$
$L$ = $\int_{{\,\ln2}}^{{\,\ln3}}$${\frac{\frac{e^{2x}+1}{e^{x}}}{\frac{e^{2x}-1}{e^{x}}}}$ $dx$
$L$ = $\int_{{\,\ln2}}^{{\,\ln3}}$${\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}}$ $dx$
$u$ = ${e^{x}-e^{-x}}$
$du$ = ${e^{x}+e^{-x}}$$dx$
$so$
$L$ = $\int$$\frac{1}{u}$ $du$
$L$ = $\ln{u}$
$L$ = $\ln({e^{x}-e^{-x}})$$|_{{\,\ln2}}^{{\,\ln3}}$
$L$ = $\ln(3-\frac{1}{3})-\ln(2-\frac{1}{2})$
$L$ = $\ln(\frac{8}{3})-\ln(\frac{3}{2})$
$L$ = $\ln(\frac{8}{3}\times\frac{2}{3})$
$L$ = $\ln(\frac{16}{9})$
