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$If f^{'} (x) = \frac{1}{- x + \sqrt{x^{2} + 1}} and f (0) = - \frac{1 + \sqrt{2}}{2} then f (1) =$

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−log⁡(2+1)

1+2

log⁡(1+2)

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Correct option is D

By rationalizing the integrand, the given integral can be written as f(x)=∫x+x2+1dx=x22+x2x2+1+log⁡x+x2+1+C Putting x=0, we have f(0)=C, so C=−1/2−1/2 and f(1)=12+122+log∣1+21+−12−12=log⁡(1+2)=−log⁡(2−1)

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If f'(x)=1−x+x2+1 and f(0)=−1+22 then f(1) =

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$If f^{'} (x) = \frac{1}{- x + \sqrt{x^{2} + 1}} and f (0) = - \frac{1 + \sqrt{2}}{2} then f (1) =$