int_0^infty frac{log(1 + x^2)}{1 + x^2} ,dx = | vedclass

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(B) Let $I = \int_0^\infty \frac{\log(1 + x^2)}{1 + x^2} \,dx$. Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \,d	heta$. When $x = 0$,$	heta

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$\pi \log 2$

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(B) Let $I = \int_0^\infty \frac{\log(1 + x^2)}{1 + x^2} \,dx$. Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \,d	heta$. When $x = 0$,$	heta = 0$,and when $x 	o \infty$,$	heta = \frac{\pi}{2}$. Thus,$I = \int_0^{\pi/2} \frac{\log(1 + 	an^2 	heta)}{1 + 	an^2 	heta} \cdot \sec^2 	heta \,d	heta$. Since $1 + 	an^2 	heta = \sec^2 	heta$,we have $I = \int_0^{\pi/2} \log(\sec^2 	heta) \,d	heta$. Using the property $\log(a^b) = b \log a$,we get $I = 2 \int_0^{\pi/2} \log(\sec 	heta) \,d	heta$. Since $\sec 	heta = \frac{1}{\cos 	heta}$,we have $\log(\sec 	heta) = -\log(\cos 	heta)$. So,$I = -2 \int_0^{\pi/2} \log(\cos 	heta) \,d	heta$. Using the standard integral result $\int_0^{\pi/2} \log(\cos 	heta) \,d	heta = -\frac{\pi}{2} \log 2$,we get: $I = -2 \left( -\frac{\pi}{2} \log 2 ight) = \pi \log 2$.

$\int_0^\infty \frac{\log(1 + x^2)}{1 + x^2} \,dx = $

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