Let u = intlimits_0^1 {frac{{ln (x + 1)}}{{{x | vedclass

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(B) For $u = \int_0^1 \frac{\ln(x+1)}{x^2+1} dx$,let $x = 	an 	heta$,then $dx = \sec^2 	heta d	heta$. When $x=0, 	heta=0$ and when $x=1, 	heta=\

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$4u + v = 0$

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(B) For $u = \int_0^1 \frac{\ln(x+1)}{x^2+1} dx$,let $x = 	an 	heta$,then $dx = \sec^2 	heta d	heta$. When $x=0, 	heta=0$ and when $x=1, 	heta=\frac{\pi}{4}$.$u = \int_0^{\pi/4} \ln(1+	an 	heta) d	heta$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$u = \int_0^{\pi/4} \ln(1+	an(\frac{\pi}{4}-	heta)) d	heta = \int_0^{\pi/4} \ln(1+\frac{1-	an 	heta}{1+	an 	heta}) d	heta = \int_0^{\pi/4} \ln(\frac{2}{1+	an 	heta}) d	heta$.$u = \int_0^{\pi/4} \ln 2 d	heta - \int_0^{\pi/4} \ln(1+	an 	heta) d	heta = \frac{\pi}{4} \ln 2 - u$.$2u = \frac{\pi}{4} \ln 2 \implies u = \frac{\pi}{8} \ln 2 \implies 4u = \frac{\pi}{2} \ln 2 \dots (1)$.For $v = \int_0^{\pi/2} \ln(\sin 2x) dx$,let $2x = t$,then $dx = \frac{1}{2} dt$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\pi$.$v = \frac{1}{2} \int_0^{\pi} \ln(\sin t) dt = \int_0^{\pi/2} \ln(\sin t) dt$.Using the standard integral $\int_0^{\pi/2} \ln(\sin t) dt = -\frac{\pi}{2} \ln 2$,we get $v = -\frac{\pi}{2} \ln 2 \dots (2)$.From $(1)$ and $(2)$,$4u = -v \implies 4u + v = 0$.

Let $u = \int\limits_0^1 {\frac{{\ln (x + 1)}}{{{x^2} + 1}}} \,dx$ and $v = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin 2x)} \,dx$,then:

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