🛠️ JEE➗ Maths

The value of the integral \({\int }_{-{\log }_{e}2}^{{\log }_{e}2}{e}^{x}\left({\log }_{e}\left({e}^{x}+\sqrt{1+{e}^{2x}…

Q1

The value of the integral \({\int }_{-{\log }_{e}2}^{{\log }_{e}2}{e}^{x}\left({\log }_{e}\left({e}^{x}+\sqrt{1+{e}^{2x}}\right)\right)dx\) is equal to

[JEE Main 2023, 11 Apr (Shift 1)]

a

\({\log }_{e}\left(\frac{2(2+\sqrt{5})}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}\)

b

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

c

\({\log }_{e}\left(\frac{(2+\sqrt{5}{)}^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}\)

d

\({\log }_{e}\left(\frac{\sqrt{2}(2+\sqrt{5}{)}^{2}}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}\)

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