The value of the integral int_{-1}^{1} log le | vedclass

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(B) Let $I = \int_{-1}^{1} f(x) \, dx$,where $f(x) = \log \left(x+\sqrt{x^{2}+1}\right)$.Check if $f(x)$ is an odd function by evaluating $f(-x)$:$f(-

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(B) Let $I = \int_{-1}^{1} f(x) \, dx$,where $f(x) = \log \left(x+\sqrt{x^{2}+1}\right)$.Check if $f(x)$ is an odd function by evaluating $f(-x)$:$f(-x) = \log \left(-x+\sqrt{(-x)^{2}+1}\right) = \log \left(\sqrt{x^{2}+1}-x\right)$.Multiply and divide by $(\sqrt{x^{2}+1}+x)$:$f(-x) = \log \left(\frac{(\sqrt{x^{2}+1}-x)(\sqrt{x^{2}+1}+x)}{\sqrt{x^{2}+1}+x}\right) = \log \left(\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}+x}\right) = \log \left(\frac{1}{\sqrt{x^{2}+1}+x}\right)$.Using the property $\log(1/a) = -\log(a)$:$f(-x) = -\log \left(x+\sqrt{x^{2}+1}\right) = -f(x)$.Since $f(-x) = -f(x)$,the function is an odd function.By the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.Therefore,$\int_{-1}^{1} \log \left(x+\sqrt{x^{2}+1}\right) \, dx = 0$.

The value of the integral $\int_{-1}^{1} \log \left(x+\sqrt{x^{2}+1}\right) \, dx$ is:

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