The integral intlimits_0^{frac{1}{2}} frac{ln | vedclass

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(C) Let $I = \int\limits_0^{1/2} \frac{\ln(1 + 2x)}{1 + (2x)^2} dx$.Substitute $2x = 	an 	heta$,then $2 dx = \sec^2 	heta d	heta$,which implies $d

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$\frac{\pi}{16} \ln 2$

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(C) Let $I = \int\limits_0^{1/2} \frac{\ln(1 + 2x)}{1 + (2x)^2} dx$.Substitute $2x = 	an 	heta$,then $2 dx = \sec^2 	heta d	heta$,which implies $dx = \frac{1}{2} \sec^2 	heta d	heta$.When $x = 0$,$	heta = 0$. When $x = 1/2$,$	heta = \pi/4$.Substituting these into the integral:$I = \int\limits_0^{\pi/4} \frac{\ln(1 + 	an 	heta)}{1 + 	an^2 	heta} \cdot \frac{1}{2} \sec^2 	heta d	heta$Since $1 + 	an^2 	heta = \sec^2 	heta$,the expression simplifies to:$I = \frac{1}{2} \int\limits_0^{\pi/4} \ln(1 + 	an 	heta) d	heta$  --- $(1)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:$I = \frac{1}{2} \int\limits_0^{\pi/4} \ln(1 + 	an(\pi/4 - 	heta)) d	heta$Using $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$:$I = \frac{1}{2} \int\limits_0^{\pi/4} \ln\left(1 + \frac{1 - 	an 	heta}{1 + 	an 	heta}ight) d	heta$$I = \frac{1}{2} \int\limits_0^{\pi/4} \ln\left(\frac{1 + 	an 	heta + 1 - 	an 	heta}{1 + 	an 	heta}ight) d	heta = \frac{1}{2} \int\limits_0^{\pi/4} \ln\left(\frac{2}{1 + 	an 	heta}ight) d	heta$$I = \frac{1}{2} \int\limits_0^{\pi/4} (\ln 2 - \ln(1 + 	an 	heta)) d	heta$$I = \frac{1}{2} \ln 2 \cdot [	heta]_0^{\pi/4} - \frac{1}{2} \int\limits_0^{\pi/4} \ln(1 + 	an 	heta) d	heta$$I = \frac{1}{2} \ln 2 \cdot \frac{\pi}{4} - I$$2I = \frac{\pi}{8} \ln 2 \implies I = \frac{\pi}{16} \ln 2$.

The integral $\int\limits_0^{\frac{1}{2}} \frac{\ln(1 + 2x)}{1 + 4x^2} dx$ equals

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