int_0^1 frac{8 log (1+x)}{1+x^2} dx = | vedclass

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(A) Let $I = \int_0^1 \frac{8 \log (1+x)}{1+x^2} dx$. Substitute $x = 	an 	heta$,so $dx = \sec^2 	heta d	heta$. When $x = 0$,$	heta = 0$,and when

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$\pi \log 2$

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(A) Let $I = \int_0^1 \frac{8 \log (1+x)}{1+x^2} dx$. Substitute $x = 	an 	heta$,so $dx = \sec^2 	heta d	heta$. When $x = 0$,$	heta = 0$,and when $x = 1$,$	heta = \frac{\pi}{4}$. Then $I = 8 \int_0^{\pi/4} \frac{\log(1+	an 	heta)}{1+	an^2 	heta} \sec^2 	heta d	heta = 8 \int_0^{\pi/4} \log(1+	an 	heta) d	heta$. Using the property $\int_0^a f(	heta) d	heta = \int_0^a f(a-	heta) d	heta$: $I = 8 \int_0^{\pi/4} \log(1+	an(\frac{\pi}{4}-	heta)) d	heta$. Since $	an(\frac{\pi}{4}-	heta) = \frac{1-	an 	heta}{1+	an 	heta}$,we have: $I = 8 \int_0^{\pi/4} \log(1 + \frac{1-	an 	heta}{1+	an 	heta}) d	heta = 8 \int_0^{\pi/4} \log(\frac{2}{1+	an 	heta}) d	heta$. $I = 8 \int_0^{\pi/4} (\log 2 - \log(1+	an 	heta)) d	heta = 8 \int_0^{\pi/4} \log 2 d	heta - I$. $2I = 8 \log 2 [	heta]_0^{\pi/4} = 8 \log 2 (\frac{\pi}{4}) = 2\pi \log 2$. Therefore,$I = \pi \log 2$.

$\int_0^1 \frac{8 \log (1+x)}{1+x^2} dx =$

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