The value of the integral \int_{-1}^{2} \log _e (x+\sqrt{x^2+1}) \dx is: [9 Apr 2024 Shift 2]

Correct Answer is : √2−√5+loge(9+4√51+√2)

\sqrt{5}-\sqrt{2}+\log _{ {\e}} ( {9+4 \sqrt{5}}/{1+\sqrt{2}})

\sqrt{2}-\sqrt{5}+\log _{ {\e}} ( {9+4 \sqrt{5}}/{1+\sqrt{2}})

\sqrt{5}-\sqrt{2}+\log _{ {\e}} ( {7+4 \sqrt{5}}/{1+\sqrt{2}})

\sqrt{2}-\sqrt{5}+\log _e ( {7+4 \sqrt{5}}/{1+\sqrt{2}})

JEE Main 9 Apr 2024 Shift 2 Solved Paper

Section: Mathematics

Question : 18 of 90

Marks: +1, -0

The value of the integral

∫

−1

loge(x+√x2+1)‌dx is:

[9 Apr 2024 Shift 2]

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