JEE Mains · Maths · STD 12 - 7.2 definite integral
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 \(\log _{e}\left(\frac{(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}\)
- D \(\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}\)
Answer & Solution
Correct Answer
(D) \(\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}\)
Step-by-step Solution
Detailed explanation
\(I=\int \limits_{-\ln 2}^{\ln 2} e^x\left(\ln \left(e^x+\sqrt{1+e^{2 x}}\right)\right) d x\) 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.…
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