The value of the integral ∫−loge2loge2ex(loge(ex+1+e2x)) dx

I=∫−loge2loge2exloge(ex+1+e2x) dxlet ex=t ⇒ ex. dx=dt x : −loge2⇒t : 12x : loge2→t :→2I=∫1/22loge(t+1+t2) dt ⇒I=t.loge(t+1+t2)|122−∫122 t1+t2dtI=2loge(2+5)−12log(1+52)−(1+t2)|122 I=loge(2+5)2+log21+5−(5−52) I=log2.(2+5)21+5−52

The value of the integral ∫−loge2loge2ex(loge(ex+1+e2x)) dx

I=∫−loge2loge2exloge(ex+1+e2x) dxlet ex=t ⇒ ex. dx=dt x : −loge2⇒t : 12x : loge2→t :→2I=∫1/22loge(t+1+t2) dt ⇒I=t.loge(t+1+t2)|122−∫122 t1+t2dtI=2loge(2+5)−12log(1+52)−(1+t2)|122 I=loge(2+5)2+log21+5−(5−52) I=log2.(2+5)21+5−52

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The value of the integral $\int_{- \log_{e} 2}^{\log_{e} 2} e^{x} (\log_{e} (e^{x} + \sqrt{1 + e^{2 x}})) dx$

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

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

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

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

answer is C.

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Detailed Solution

$I = \int_{- {\log_{e}}^{2}}^{{\log_{e}}^{2}} e^{x} \log_{e} (e^{x} + \sqrt{1 + e^{2 x}}) dx$
$let e^{x} = t \Rightarrow e^{x} . dx = dt$ $\begin{array}{l} x : - {\log_{e}}^{2} \Rightarrow t : \frac{1}{2} \\ x : {\log_{e}}^{2} \to t : \to 2 \end{array}$
$I = \int_{1 / 2}^{2} \log_{e} (t + \sqrt{1 + t^{2}}) dt$ $\Rightarrow I = {t . \log_{e} (t + \sqrt{1 + t^{2}}) |}_{\frac{1}{2}}^{2} - \int_{\frac{1}{2}}^{2} \frac{t}{\sqrt{1 + t^{2}}} dt$
$I = {2 {\log_{e}}^{(2 + \sqrt{5})} - \frac{1}{2} \log (\frac{1 + \sqrt{5}}{2}) - (\sqrt{1 + t^{2}}) |}_{\frac{1}{2}}^{2} {I = \log_{e}}^{{(2 + \sqrt{5})}^{2}} + \log \frac{\sqrt{2}}{\sqrt{1 + \sqrt{5}}} - (\sqrt{5} - \frac{\sqrt{5}}{2}) I = \log \frac{\sqrt{2} . {(2 + \sqrt{5})}^{2}}{\sqrt{1 + \sqrt{5}}} - \frac{\sqrt{5}}{2}$