MathematicsMedium210×since 2002Q3348The value of ∫018log(1+x)1+x2dx\int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx0∫11+x28log(1+x)dx isAπ8log2{\pi \over 8}\log 28πlog2Bπ2log2{\pi \over 2}\log 22πlog2Clog2\log 2log2Dπlog2\pi \log 2πlog2Check answerSkip