Evaluate-0-2-e-x-e-1-x-e-x-1-dx-

Question Number 174548 by mnjuly1970 last updated on 03/Aug/22

E v a l u a t e .

Ω = \int_{0}^{2} \frac{e^{x}}{e^{1 - x} + e^{x - 1}} d x = ?

Answered by CElcedricjunior last updated on 03/Aug/22

Ω = \int_{0}^{2} \frac{e^{x}}{e^{1 - x} + e^{x - 1}} dx = \int_{0}^{2} \frac{e^{x - 1} \times e^{x} d x}{1 + e^{x - 1} \times e^{x - 1}}

= \int_{0}^{2} \frac{e^{2 x - 1}}{1 + e^{2 x - 2}} d x = \frac{e}{2} \int_{0}^{2} \frac{2 e^{2 x - 2}}{1 + e^{2 x - 2}} d x

Ω = \frac{e}{2} {[\ln (1 + e^{2 x - 2})]}_{0}^{2} = \frac{e}{2} [\ln (1 + e^{2}) - \ln (1 + \overset{- 2}{e})]

= \frac{e}{2} {lne}^{2}

Ω = e

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