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

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(B) 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$; w

Vedclass · Accepted Answer

$\pi \log 2$

Vedclass · Answer

(B) 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$; when $x = 1$, $	heta = \frac{\pi}{4}$. $I = \int_0^{\frac{\pi}{4}} \frac{8 \log (1+	an 	heta)}{1+	an^2 	heta} \cdot \sec^2 	heta \,d	heta = \int_0^{\frac{\pi}{4}} 8 \log (1+	an 	heta) \,d	heta \quad \dots(i)$ Using the property $\int_0^a f(	heta) \,d	heta = \int_0^a f(a-	heta) \,d	heta$: $I = \int_0^{\frac{\pi}{4}} 8 \log \left(1 + 	an\left(\frac{\pi}{4} - 	hetaight)ight) \,d	heta$ Since $	an\left(\frac{\pi}{4} - 	hetaight) = \frac{1-	an 	heta}{1+	an 	heta}$, we have: $I = \int_0^{\frac{\pi}{4}} 8 \log \left(1 + \frac{1-	an 	heta}{1+	an 	heta}ight) \,d	heta = \int_0^{\frac{\pi}{4}} 8 \log \left(\frac{2}{1+	an 	heta}ight) \,d	heta \quad \dots(ii)$ Adding $(i)$ and $(ii)$: $2I = \int_0^{\frac{\pi}{4}} 8 \left[ \log(1+	an 	heta) + \log\left(\frac{2}{1+	an 	heta}ight) ight] \,d	heta$ $2I = 8 \int_0^{\frac{\pi}{4}} \log \left( (1+	an 	heta) \cdot \frac{2}{1+	an 	heta} ight) \,d	heta$ $2I = 8 \int_0^{\frac{\pi}{4}} \log 2 \,d	heta = 8 \log 2 [	heta]_0^{\frac{\pi}{4}} = 8 \log 2 \cdot \frac{\pi}{4} = 2\pi \log 2$. Therefore, $I = \pi \log 2$.

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

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