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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
The value of the integral $\int \limits_{-\log _{ e } 2}^{\log _e 2} e^x\left(\log _0\left(e^x+\sqrt{1+e^{2 x}}\right)\right) d x$ is equal to
  • 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
    $\log _{e}\left(\frac{(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}$
  • ✓
    $\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}$
✓

Answer

Correct option: D.
$\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}$
d
$I=\int \limits_{-\ln 2}^{\ln 2} e^x\left(\ln \left(e^x+\sqrt{1+e^{2 x}}\right)\right) d x$

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

$I=\int \limits_{1 / 2}^2 \ln \left(t+\sqrt{1+t^2}\right) d t$

Applying integration by parts.

$=\left[t \ln \left( t +\sqrt{1+ t ^2}\right)\right]_{\frac{1}{2}}^2-\int \limits_{1 / 2}^2 \frac{ t }{ t +\sqrt{1+ t ^2}}\left(1+\frac{2 t }{2 \sqrt{1+ t ^2}}\right) d t$

$=2 \ln (2+\sqrt{5})-\frac{1}{2} \ln \left(\frac{1+\sqrt{5}}{2}\right)-\int \limits_{1 / 2}^2 \frac{ t }{\sqrt{1+ t ^2}} dt$

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

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

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