The value of the integral - log e 2 log e 2 e x ( log e ( e x + 1 + e 2 x ) ) d x is equal to [2023]

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Solution

Q.
The value of the integral $\int_{- \log_{e} 2}^{\log_{e} 2} e^{x} (\log_{e} (e^{x} + \sqrt{1 + e^{2 x}})) d x$ is equal to [2023]

1 $\log_{e} (\frac{2 (2 + \sqrt{5})}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$

2 $\log_{e} (\frac{\sqrt{2} {(3 - \sqrt{5})}^{2}}{\sqrt{1 + \sqrt{5}}}) + \frac{\sqrt{5}}{2}$

3 $\log_{e} (\frac{{(2 + \sqrt{5})}^{2}}{\sqrt{1 + \sqrt{5}}}) + \frac{\sqrt{5}}{2}$

4 $\log_{e} (\frac{\sqrt{2} {(2 + \sqrt{5})}^{2}}{\sqrt{1 + \sqrt{5}}}) - \frac{\sqrt{5}}{2}$

Ans.
(4)

$Let I = \int_{- \log_{e} 2}^{\log_{e} 2} e^{x} (\ln (e^{x} + \sqrt{1 + e^{2 x}})) d x$

$Put e^{x} = t \Rightarrow e^{x} d x = d t$

$When x = - \log_{e} 2, t = \frac{1}{2}; when x = \log_{e} 2, t = 2$

$∴$ $I = \int_{1 / 2}^{2} \ln (t + \sqrt{1 + t^{2}}) d t$

Applying integration by parts, we get

$I = {[t \ln (t + \sqrt{1 + t^{2}})]}_{1 / 2}^{2} - \int_{1 / 2}^{2} \frac{t}{t + \sqrt{1 + t^{2}}} (1 + \frac{2 t}{2 \sqrt{1 + t^{2}}}) d t$

$= 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1 + \sqrt{5}}{2}) - \int_{1 / 2}^{2} \frac{t}{\sqrt{1 + t^{2}}} d t$

$= 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1 + \sqrt{5}}{2}) - {[\sqrt{1 + t^{2}}]}_{1 / 2}^{2}$

$= 2 \ln (2 + \sqrt{5}) - \frac{1}{2} \ln (\frac{1 + \sqrt{5}}{2}) - \frac{\sqrt{5}}{2}$ $= \ln (\frac{{(2 + \sqrt{5})}^{2}}{{(\frac{\sqrt{5} + 1}{2})}^{\frac{1}{2}}}) - \frac{\sqrt{5}}{2}$

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