Find the value of int_{0}^{1} frac{log (1+x)} | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$.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}$.Thus,$I = \int_{0}^{\pi/4} \frac{\log (1+	an 	heta)}{1+	an^{2} 	heta} (\sec^{2} 	heta) d 	heta = \int_{0}^{\pi/4} \log (1+	an 	heta) d 	heta \quad ...(i)$Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we have:$I = \int_{0}^{\pi/4} \log \left[1+	an \left(\frac{\pi}{4}-	hetaight)ight] d 	heta$$I = \int_{0}^{\pi/4} \log \left[1+\frac{1-	an 	heta}{1+	an 	heta}ight] d 	heta = \int_{0}^{\pi/4} \log \left(\frac{2}{1+	an 	heta}ight) d 	heta$$I = \int_{0}^{\pi/4} [\log 2 - \log (1+	an 	heta)] d 	heta \quad ...(ii)$Adding $(i)$ and $(ii)$:$2I = \int_{0}^{\pi/4} \log 2 d 	heta = \log 2 [	heta]_{0}^{\pi/4} = \frac{\pi}{4} \log 2$.Therefore,$I = \frac{\pi}{8} \log 2$.

Find the value of $\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$.

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