Question:
The value of the integral $\int_{-1}^{1} \log \left(x+\sqrt{x^{2}+1}\right) d x$ is:
Correct Option: , 2
Solution:
Let $I=\int_{-1}^{1} \log \left(x+\sqrt{x^{2}+1}\right) d x$
$\because \log \left(x+\sqrt{x^{2}+1}\right)$ is an odd function
$\therefore I=0$
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