Let u = int_0^1 frac{ln(x + 1)}{x^2 + 1} , dx | vedclass

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

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$4u + v = 0$

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(B) Given $u = \int_0^1 \frac{\ln(x + 1)}{x^2 + 1} \, dx$. Let $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$. When $x=0, 	heta=0$; when $x=1, 	heta=\frac{\pi}{4}$.$u = \int_0^{\frac{\pi}{4}} \ln(1 + 	an 	heta) \, d	heta$.Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$u = \int_0^{\frac{\pi}{4}} \ln(1 + 	an(\frac{\pi}{4} - 	heta)) \, d	heta = \int_0^{\frac{\pi}{4}} \ln(1 + \frac{1 - 	an 	heta}{1 + 	an 	heta}) \, d	heta = \int_0^{\frac{\pi}{4}} \ln(\frac{2}{1 + 	an 	heta}) \, d	heta$.$u = \int_0^{\frac{\pi}{4}} (\ln 2 - \ln(1 + 	an 	heta)) \, d	heta = \frac{\pi}{4} \ln 2 - u$.$2u = \frac{\pi}{4} \ln 2 \implies u = \frac{\pi}{8} \ln 2$.Now,$v = \int_0^{\frac{\pi}{2}} \ln(\sin 2x) \, dx$. Let $2x = t$,then $dx = \frac{1}{2} dt$. When $x=0, t=0$; when $x=\frac{\pi}{2}, t=\pi$.$v = \frac{1}{2} \int_0^{\pi} \ln(\sin t) \, dt = \frac{1}{2} \cdot 2 \int_0^{\frac{\pi}{2}} \ln(\sin t) \, dt = \int_0^{\frac{\pi}{2}} \ln(\sin t) \, dt$.Using the standard result $\int_0^{\frac{\pi}{2}} \ln(\sin t) \, dt = -\frac{\pi}{2} \ln 2$,we get $v = -\frac{\pi}{2} \ln 2$.Comparing $u$ and $v$: $4u = 4(\frac{\pi}{8} \ln 2) = \frac{\pi}{2} \ln 2 = -v$.Therefore,$4u + v = 0$.

Let $u = \int_0^1 \frac{\ln(x + 1)}{x^2 + 1} \, dx$ and $v = \int_0^{\frac{\pi}{2}} \ln(\sin 2x) \, dx$,then:

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