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$\int_{- 1}^{1} \log (x + \sqrt{x^{2} + 1}) dx$ is equal to

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Correct option is A

Let f(x)=log⁡x+1+x2 and replacing x by we getf(−x)=log⁡1+x2−x=log⁡1+x2−x1+x2+x1+x2+x=log⁡1+x2−x21+x2+x=log⁡1−log⁡1+x2+x=−log⁡1+x2+x⇒f(−x)=−f(x)Hence , f (x) is an odd function.∴ ∫−11 log⁡x+1+x2dx=0

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∫−11 log⁡x+x2+1dx is equal to

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$\int_{- 1}^{1} \log (x + \sqrt{x^{2} + 1}) dx$ is equal to