intlimits_0^1 {x,ln left( {1 + frac{x}{2}} ri | vedclass

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Question

(A) Let $I = \int\limits_0^1 {x \ln \left( {\frac{{x + 2}}{2}} \right) dx} = \int\limits_0^1 {x (\ln (x + 2) - \ln 2) dx}$.$= \int\limits_0^1 {x \ln (

Vedclass · Accepted Answer

$\frac{3}{4}\left( {1 - 2\ln \frac{3}{2}} \right)$

Vedclass · Answer

(A) Let $I = \int\limits_0^1 {x \ln \left( {\frac{{x + 2}}{2}} \right) dx} = \int\limits_0^1 {x (\ln (x + 2) - \ln 2) dx}$.$= \int\limits_0^1 {x \ln (x + 2) dx} - \ln 2 \int\limits_0^1 {x dx}$.Using integration by parts for the first term: $\left[ \frac{x^2}{2} \ln (x + 2) \right]_0^1 - \int\limits_0^1 {\frac{x^2}{2(x + 2)} dx} - \frac{\ln 2}{2}$.$= \frac{1}{2} \ln 3 - \frac{1}{2} \int\limits_0^1 {\frac{x^2 - 4 + 4}{x + 2} dx} - \frac{\ln 2}{2} = \frac{1}{2} \ln \frac{3}{2} - \frac{1}{2} \int\limits_0^1 {(x - 2 + \frac{4}{x + 2}) dx}$.$= \frac{1}{2} \ln \frac{3}{2} - \frac{1}{2} \left[ \frac{x^2}{2} - 2x + 4 \ln (x + 2) \right]_0^1$.$= \frac{1}{2} \ln \frac{3}{2} - \frac{1}{2} [(\frac{1}{2} - 2 + 4 \ln 3) - (0 - 0 + 4 \ln 2)]$.$= \frac{1}{2} \ln \frac{3}{2} - \frac{1}{2} [-\frac{3}{2} + 4 \ln \frac{3}{2}] = \frac{1}{2} \ln \frac{3}{2} + \frac{3}{4} - 2 \ln \frac{3}{2} = \frac{3}{4} - \frac{3}{2} \ln \frac{3}{2} = \frac{3}{4} (1 - 2 \ln \frac{3}{2})$.Thus,the correct option is $A$.

$\int\limits_0^1 {x\,\ln \left( {1 + \frac{x}{2}} \right)\,dx} =$

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