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

Question
Integrals
Evaluate the following integrals.
$$\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx$$

2020-11-02
$$\text{To evaluate:}$$
$$\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx$$
$$\text{We have,}$$
$$(e^x-e^{-x})^2=e^{2x}-2+e^{-2x}$$
$$\text{Thus,}$$
$$\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx=\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{(e^x-e^{-x})^2}dx$$
$$\text{Substitute, }e^x-e^{-x}=u\Rightarrow(e^x+e^{-x})dx=du$$
$$\text{Therefore, becomes}$$
$$\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx=\int_{\frac{3}{2}}^{\frac{8}{3}}\frac{1}{(u)^2}du$$
$$\int_{\frac{3}{2}}^{\frac{8}{3}}u^{-2}du$$
$$=[\frac{u^{-1}}{-1}]_{\frac{3}{2}}^{\frac{8}{3}}$$
$$=[\frac{-1}{u}]_{\frac{3}{2}}^{\frac{8}{3}}$$
$$=-\frac{3}{8}+\frac{2}{3}$$
$$=\frac{-9+16}{24}$$
$$=\frac{7}{24}$$
$$\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx=\frac{7}{24}$$

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