int_{-1}^{1} log(x + sqrt{x^2 + 1}) , dx = | vedclass

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(A) Let $f(x) = \log(x + \sqrt{1 + x^2})$.Now,$f(-x) = \log(-x + \sqrt{1 + (-x)^2}) = \log(\sqrt{1 + x^2} - x)$.Multiplying and dividing by $(\sqrt{1

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(A) Let $f(x) = \log(x + \sqrt{1 + x^2})$.Now,$f(-x) = \log(-x + \sqrt{1 + (-x)^2}) = \log(\sqrt{1 + x^2} - x)$.Multiplying and dividing by $(\sqrt{1 + x^2} + x)$,we get:$f(-x) = \log\left(\frac{(\sqrt{1 + x^2} - x)(\sqrt{1 + x^2} + x)}{\sqrt{1 + x^2} + x}\right)$$= \log\left(\frac{(1 + x^2) - x^2}{\sqrt{1 + x^2} + x}\right)$$= \log\left(\frac{1}{\sqrt{1 + x^2} + x}\right)$$= \log(1) - \log(\sqrt{1 + x^2} + x) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.By the property of definite integrals,$\int_{-a}^{a} f(x) \, dx = 0$ if $f(x)$ is an odd function.Therefore,$\int_{-1}^{1} \log(x + \sqrt{1 + x^2}) \, dx = 0$.

$\int_{-1}^{1} \log(x + \sqrt{x^2 + 1}) \, dx = $

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