MathematicsMedium210×since 2002Q3363A value of α\alphaα such that ∫αα+1dx(x+α)(x+α+1)=loge(98)\int\limits_\alpha ^{\alpha + 1} {{{dx} \over {\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}} = {\log _e}\left( {{9 \over 8}} \right)α∫α+1(x+α)(x+α+1)dx=loge(89) is :A2B- 2C12{1 \over 2}21D−12-{1 \over 2}−21Check answerSkip