int_{0}^{1} frac{8 log(1+x)}{1+x^{2}} dx = | vedclass

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(D) 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 wh

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$\pi \log 2$

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(D) 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 $x=1, 	heta=\frac{\pi}{4}$. $I = 8 \int_{0}^{\pi/4} \frac{\log(1+	an 	heta)}{1+	an^{2} 	heta} \sec^{2} 	heta d	heta = 8 \int_{0}^{\pi/4} \log(1+	an 	heta) d	heta$ ... $(i)$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$: $I = 8 \int_{0}^{\pi/4} \log(1+	an(\frac{\pi}{4}-	heta)) d	heta$. Since $	an(\frac{\pi}{4}-	heta) = \frac{1-	an 	heta}{1+	an 	heta}$,we have: $I = 8 \int_{0}^{\pi/4} \log(1 + \frac{1-	an 	heta}{1+	an 	heta}) d	heta = 8 \int_{0}^{\pi/4} \log(\frac{2}{1+	an 	heta}) d	heta$. $I = 8 \int_{0}^{\pi/4} (\log 2 - \log(1+	an 	heta)) d	heta$. $I = 8 \log 2 [	heta]_{0}^{\pi/4} - 8 \int_{0}^{\pi/4} \log(1+	an 	heta) d	heta$. $I = 8 \log 2 (\frac{\pi}{4}) - I$. $2I = 2\pi \log 2 \Rightarrow I = \pi \log 2$.

$\int_{0}^{1} \frac{8 \log(1+x)}{1+x^{2}} dx = $

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