The value of the integral {\int}↙\stackrel{\log _e 2}{-\log _e 2} e^x (\log _e (e^x+\sqrt{1+e^{2 x}) ) is equal to : [11-Apr-2023 shift 1]

Correct Answer is : loge(√2(2+√5)2√1+√5)−√52

\log _{ {\e}} ( {(2+\sqrt{5})^2}/\sqrt{1+{\sqrt{5}}}) + {\sqrt{5}}/{2}

\log _{ {\e}} ( {2(2+\sqrt{5})^2}/\sqrt{1+{\sqrt{5}}}) - {\sqrt{5}}/{2}

\log _{ {\e}} ( {\sqrt{2}(3-\sqrt{5})^2}/\sqrt{1+{\sqrt{5}}}) + {\sqrt{5}}/{2}

\log _{ {\e}} ( {\sqrt{2}(2+\sqrt{5})^2}/\sqrt{1+{\sqrt{5}}}) - {\sqrt{5}}/{2}

Definite Integrals Part 3

Section: Mathematics

Question : 91 of 115

Marks: +1, -0

The value of the integral

loge2

∫

−loge2

ex(loge(ex+√1+e2x)) is equal to :

[11-Apr-2023 shift 1]

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