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

Question
Integrals
asked 2020-11-01
Evaluate the following integrals.
\(\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx\)

Answers (1)

2020-11-02
\(\text{To evaluate:}\)
\(\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx\)
\(\text{We have,}\)
\((e^x-e^{-x})^2=e^{2x}-2+e^{-2x}\)
\(\text{Thus,}\)
\(\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx=\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{(e^x-e^{-x})^2}dx\)
\(\text{Substitute, }e^x-e^{-x}=u\Rightarrow(e^x+e^{-x})dx=du\)
\(\text{Therefore, becomes}\)
\(\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx=\int_{\frac{3}{2}}^{\frac{8}{3}}\frac{1}{(u)^2}du\)
\(\int_{\frac{3}{2}}^{\frac{8}{3}}u^{-2}du\)
\(=[\frac{u^{-1}}{-1}]_{\frac{3}{2}}^{\frac{8}{3}}\)
\(=[\frac{-1}{u}]_{\frac{3}{2}}^{\frac{8}{3}}\)
\(=-\frac{3}{8}+\frac{2}{3}\)
\(=\frac{-9+16}{24}\)
\(=\frac{7}{24}\)
\(\int_{\ln2}^{\ln3}\frac{e^x+e^{-x}}{e^{2x}-2+e^{-2x}}dx=\frac{7}{24}\)
0

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