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

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Question

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

Vedclass · Accepted Answer

$\frac{\pi}{8} \log _e 2$

Vedclass · Answer

(D) Let $I = \int_0^1 \frac{\log _e(1+x)}{1+x^2} d x$. Substitute $x = 	an 	heta$,then $d x = \sec^2 	heta d 	heta$. When $x = 0$,$	heta = 0$,and when $x = 1$,$	heta = \frac{\pi}{4}$. So,$I = \int_0^{\frac{\pi}{4}} \frac{\log _e(1+	an 	heta)}{1+	an^2 	heta} \sec^2 	heta d 	heta = \int_0^{\frac{\pi}{4}} \log _e(1+	an 	heta) d 	heta$ ... $(i)$. Using the property $\int_0^a f(	heta) d 	heta = \int_0^a f(a-	heta) d 	heta$,we get: $I = \int_0^{\frac{\pi}{4}} \log _e(1+	an(\frac{\pi}{4}-	heta)) d 	heta$. Since $	an(\frac{\pi}{4}-	heta) = \frac{1-	an 	heta}{1+	an 	heta}$,we have: $I = \int_0^{\frac{\pi}{4}} \log _e(1 + \frac{1-	an 	heta}{1+	an 	heta}) d 	heta = \int_0^{\frac{\pi}{4}} \log _e(\frac{2}{1+	an 	heta}) d 	heta$. $I = \int_0^{\frac{\pi}{4}} \log _e 2 d 	heta - \int_0^{\frac{\pi}{4}} \log _e(1+	an 	heta) d 	heta$. $I = \frac{\pi}{4} \log _e 2 - I$. $2I = \frac{\pi}{4} \log _e 2 \implies I = \frac{\pi}{8} \log _e 2$. Thus,option $(D)$ is correct.

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

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