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(A) Let $I = \int_{0}^{\infty} \frac{x \ln x}{(1 + x^2)^2} \, dx$.Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.As $x 	o 0$,$	he

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(A) Let $I = \int_{0}^{\infty} \frac{x \ln x}{(1 + x^2)^2} \, dx$.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 \frac{\pi}{2}$.Substituting these into the integral:$I = \int_{0}^{\pi/2} \frac{	an 	heta \ln(	an 	heta)}{(1 + 	an^2 	heta)^2} \sec^2 	heta \, d	heta$Since $1 + 	an^2 	heta = \sec^2 	heta$,we have:$I = \int_{0}^{\pi/2} \frac{	an 	heta \ln(	an 	heta)}{\sec^4 	heta} \sec^2 	heta \, d	heta$$I = \int_{0}^{\pi/2} \frac{	an 	heta}{\sec^2 	heta} \ln(	an 	heta) \, d	heta$$I = \int_{0}^{\pi/2} \sin 	heta \cos 	heta \ln(	an 	heta) \, d	heta$Using $\sin 	heta \cos 	heta = \frac{1}{2} \sin 2	heta$:$I = \frac{1}{2} \int_{0}^{\pi/2} \sin 2	heta \ln(	an 	heta) \, d	heta$Let $J = \int_{0}^{\pi/2} \sin 2	heta \ln(	an 	heta) \, d	heta$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$:$J = \int_{0}^{\pi/2} \sin(2(\frac{\pi}{2} - 	heta)) \ln(	an(\frac{\pi}{2} - 	heta)) \, d	heta$$J = \int_{0}^{\pi/2} \sin(\pi - 2	heta) \ln(\cot 	heta) \, d	heta$$J = \int_{0}^{\pi/2} \sin 2	heta \ln(\cot 	heta) \, d	heta$$J = \int_{0}^{\pi/2} \sin 2	heta (-\ln(	an 	heta)) \, d	heta = -J$Thus,$2J = 0$,which implies $J = 0$. Therefore,$I = 0$.

$\int_{0}^{\infty} \frac{x \ln x}{(1 + x^2)^2} \, dx$ is equal to

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