If f(x) = int frac{dx}{x^2+2} and f(sqrt{2}) | vedclass

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(C) Given $f(x) = \int \frac{dx}{x^2+(\sqrt{2})^2}$.Using the standard integral formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a})

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$\frac{-\pi}{4 \sqrt{2}}$

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(C) Given $f(x) = \int \frac{dx}{x^2+(\sqrt{2})^2}$.Using the standard integral formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$f(x) = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{x}{\sqrt{2}}) + C$.Given $f(\sqrt{2}) = 0$,substitute $x = \sqrt{2}$:$0 = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{\sqrt{2}}{\sqrt{2}}) + C$$0 = \frac{1}{\sqrt{2}} 	an^{-1}(1) + C$$0 = \frac{1}{\sqrt{2}} (\frac{\pi}{4}) + C$$C = -\frac{\pi}{4\sqrt{2}}$.Now,find $f(0)$:$f(0) = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{0}{\sqrt{2}}) - \frac{\pi}{4\sqrt{2}}$$f(0) = \frac{1}{\sqrt{2}} (0) - \frac{\pi}{4\sqrt{2}} = -\frac{\pi}{4\sqrt{2}}$.

If $f(x) = \int \frac{dx}{x^2+2}$ and $f(\sqrt{2}) = 0$,then $f(0) =$

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