# Evaluate the following integrals. int_{ln2}^{ln3}frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx

Evaluate the following integrals.
${\int }_{\mathrm{ln}2}^{\mathrm{ln}3}\frac{{e}^{x}+{e}^{-x}}{{e}^{2x}-2+{e}^{-2x}}dx$
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Asma Vang
$\text{To evaluate:}$
${\int }_{\mathrm{ln}2}^{\mathrm{ln}3}\frac{{e}^{x}+{e}^{-x}}{{e}^{2x}-2+{e}^{-2x}}dx$
$\text{We have,}$
$\left({e}^{x}-{e}^{-x}{\right)}^{2}={e}^{2x}-2+{e}^{-2x}$
$\text{Thus,}$
${\int }_{\mathrm{ln}2}^{\mathrm{ln}3}\frac{{e}^{x}+{e}^{-x}}{{e}^{2x}-2+{e}^{-2x}}dx={\int }_{\mathrm{ln}2}^{\mathrm{ln}3}\frac{{e}^{x}+{e}^{-x}}{\left({e}^{x}-{e}^{-x}{\right)}^{2}}dx$

$\text{Therefore, becomes}$
${\int }_{\mathrm{ln}2}^{\mathrm{ln}3}\frac{{e}^{x}+{e}^{-x}}{{e}^{2x}-2+{e}^{-2x}}dx={\int }_{\frac{3}{2}}^{\frac{8}{3}}\frac{1}{\left(u{\right)}^{2}}du$
${\int }_{\frac{3}{2}}^{\frac{8}{3}}{u}^{-2}du$
$=\left[\frac{{u}^{-1}}{-1}{\right]}_{\frac{3}{2}}^{\frac{8}{3}}$
$=\left[\frac{-1}{u}{\right]}_{\frac{3}{2}}^{\frac{8}{3}}$
$=-\frac{3}{8}+\frac{2}{3}$
$=\frac{-9+16}{24}$
$=\frac{7}{24}$
${\int }_{\mathrm{ln}2}^{\mathrm{ln}3}\frac{{e}^{x}+{e}^{-x}}{{e}^{2x}-2+{e}^{-2x}}dx=\frac{7}{24}$