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(A) Let $I = \int_0^\infty \log \left( x + \frac{1}{x} ight) \frac{dx}{1 + x^2}$.Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.A

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$\pi \log 2$

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(A) Let $I = \int_0^\infty \log \left( x + \frac{1}{x} ight) \frac{dx}{1 + x^2}$.Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.As $x 	o 0, 	heta 	o 0$ and as $x 	o \infty, 	heta 	o \pi / 2$.$I = \int_0^{\pi / 2} \log (	an 	heta + \cot 	heta) \frac{\sec^2 	heta}{1 + 	an^2 	heta} \, d	heta = \int_0^{\pi / 2} \log \left( \frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} ight) d	heta$.$I = \int_0^{\pi / 2} \log \left( \frac{1}{\sin 	heta \cos 	heta} ight) d	heta = - \int_0^{\pi / 2} \log (\sin 	heta \cos 	heta) \, d	heta$.$I = - \int_0^{\pi / 2} (\log \sin 	heta + \log \cos 	heta) \, d	heta$.Using the property $\int_0^{\pi / 2} \log \sin 	heta \, d	heta = \int_0^{\pi / 2} \log \cos 	heta \, d	heta = -\frac{\pi}{2} \log 2$.$I = - \left( -\frac{\pi}{2} \log 2 - \frac{\pi}{2} \log 2 ight) = - (-\pi \log 2) = \pi \log 2$.

$\int_{0}^{\infty} \log \left( x + \frac{1}{x} \right) \frac{dx}{1 + x^2}$ is equal to

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