Given-log-5-7-a-2-log-7-5-a-2-Find-the-value-a-e-1-ln-x-x-x-x-x-dx-

Question Number 136883 by bramlexs22 last updated on 27/Mar/21

Given \log_{5} (7^{a} - 2) = \log_{7} (5^{a} + 2)

. Find the value \int_{a}^{e} \frac{1 + \ln (x)}{x^{x} + x^{- x}} dx .

Answered by Olaf last updated on 27/Mar/21

\log_{5} (7^{a} - 2) = \log_{7} (5^{a} + 2)

\Rightarrow a = 1 (trivial solution)

Ω = \int_{1}^{e} \frac{1 + \ln x}{x^{x} + x^{- x}} d x

Ω = \int_{1}^{e} \frac{1 + \ln x}{e^{x \ln x} + e^{- x \ln x}} d x

Ω = \int_{1}^{e} \frac{d (x \ln x)}{e^{x \ln x} + e^{- x \ln x}}

Ω = \int_{1}^{e} \frac{e^{x \ln x}}{1 + e^{2 x \ln x}} d (x \ln x)

Ω = {[\arctan (e^{x \ln x})]}_{1}^{e}

Ω = \arctan (e^{e}) - \frac{π}{4}

Answered by liberty last updated on 27/Mar/21

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