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Question
A value of $\alpha $ such that
$\int\limits_\alpha ^{\alpha + 1} {{{dx} \over {\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}} = {\log _e}\left( {{9 \over 8}} \right)$ is :
2
- 2
${1 \over 2}$
$-{1 \over 2}$

Solution

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