The value of the definite integral intlimits_ | vedclass

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Question

(D) Let $I = \int\limits_0^{1/2} \frac{\ln(1 + 2x)}{1 + 4x^2} \,dx$.Substitute $2x = 	an 	heta$,then $2 \,dx = \sec^2 	heta \,d	heta$,so $dx = \fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int\limits_0^{1/2} \frac{\ln(1 + 2x)}{1 + 4x^2} \,dx$.Substitute $2x = 	an 	heta$,then $2 \,dx = \sec^2 	heta \,d	heta$,so $dx = \frac{1}{2} \sec^2 	heta \,d	heta$.When $x = 0$,$	heta = 0$. When $x = 1/2$,$	an 	heta = 1$,so $	heta = \pi/4$.The integral becomes $I = \int\limits_0^{\pi/4} \frac{\ln(1 + 	an 	heta)}{1 + 	an^2 	heta} \cdot \frac{1}{2} \sec^2 	heta \,d	heta$.Since $1 + 	an^2 	heta = \sec^2 	heta$,we have $I = \frac{1}{2} \int\limits_0^{\pi/4} \ln(1 + 	an 	heta) \,d	heta$.Using the property $\int\limits_0^a f(x) \,dx = \int\limits_0^a f(a-x) \,dx$,we have $I = \frac{1}{2} \int\limits_0^{\pi/4} \ln(1 + 	an(\pi/4 - 	heta)) \,d	heta$.Since $	an(\pi/4 - 	heta) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$,we get $1 + 	an(\pi/4 - 	heta) = 1 + \frac{1 - 	an 	heta}{1 + 	an 	heta} = \frac{2}{1 + 	an 	heta}$.Thus,$I = \frac{1}{2} \int\limits_0^{\pi/4} (\ln 2 - \ln(1 + 	an 	heta)) \,d	heta = \frac{1}{2} [	heta \ln 2]_0^{\pi/4} - I$.$2I = \frac{1}{2} \cdot \frac{\pi}{4} \ln 2 = \frac{\pi \ln 2}{8}$.Therefore,$I = \frac{\pi \ln 2}{16}$.

The value of the definite integral $\int\limits_0^{\frac{1}{2}} \frac{\ln(1 + 2x)}{1 + 4x^2} \,dx$ is equal to:

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