We have an answer to "Evaluate the following integrals. ∫ln⁡2ln⁡3ex+e−xe2x−2+e−2xdx"

nagasenaz

Answered question

2020-11-01

Evaluate the following integrals.

\int_{\ln 2}^{\ln 3} \frac{e^{x} + e^{- x}}{e^{2 x} - 2 + e^{- 2 x}} d x

Asma Vang

Skilled2020-11-02Added 93 answers

To evaluate:

\int_{\ln 2}^{\ln 3} \frac{e^{x} + e^{- x}}{e^{2 x} - 2 + e^{- 2 x}} d x

We have,

(e^{x} - e^{- x})^{2} = e^{2 x} - 2 + e^{- 2 x}

Thus,

\int_{\ln 2}^{\ln 3} \frac{e^{x} + e^{- x}}{e^{2 x} - 2 + e^{- 2 x}} d x = \int_{\ln 2}^{\ln 3} \frac{e^{x} + e^{- x}}{(e^{x} - e^{- x})^{2}} d x

Substitute, e^{x} - e^{- x} = u \Rightarrow (e^{x} + e^{- x}) d x = d u

Therefore, becomes

\int_{\ln 2}^{\ln 3} \frac{e^{x} + e^{- x}}{e^{2 x} - 2 + e^{- 2 x}} d x = \int_{\frac{3}{2}}^{\frac{8}{3}} \frac{1}{(u)^{2}} d u

\int_{\frac{3}{2}}^{\frac{8}{3}} u^{- 2} d u

= [\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_{\ln 2}^{\ln 3} \frac{e^{x} + e^{- x}}{e^{2 x} - 2 + e^{- 2 x}} d x = \frac{7}{24}

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