Official solution
\[I = \int\limits_\alpha ^{\alpha + 1} {{{dx} \over {(x + \alpha )(x + \alpha + 1)}}} \]
Let x + \[\alpha \] = t and dx = dt
\[I = \int\limits_{2\alpha }^{2\alpha + 1} {{{dt} \over {t(t + 1)}}} = \int\limits_{2\alpha }^{2\alpha + 1} {\left( {{1 \over t} - {1 \over {t + 1}}} \right)dx} \]
\[ \Rightarrow \left( {\ln t - \ln (t + 1)} \right)|_{2\alpha }^{2\alpha + 1} = \ln \left( {{t \over {t + 1}}} \right)|_{2\alpha }^{2\alpha + 1}\]
\[ \Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}}} \right) - \ln \left( {{{2\alpha } \over {2\alpha + 1}}} \right)\]
\[ \Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}} \times {{2\alpha + 1} \over {2\alpha }}} \right) = \ln \left( {{9 \over 8}} \right)\]
\[ \Rightarrow {{{{\left( {2\alpha + 1} \right)}^2}} \over {\left( {2\alpha + 2} \right)}} = {{9\alpha } \over 4}\]
\[4(4{\alpha ^2} + 1 + 4\alpha ) = 18{\alpha ^2} + 18\alpha \]
\[ \Rightarrow 2{\alpha ^2} - 2\alpha - 4 = 0\]
\[ \Rightarrow {\alpha ^2} + \alpha - 2 = 0\]
\[ \Rightarrow (\alpha + 2)(\alpha - 1) = 0\]
\[\alpha \] = 1 or \[\alpha \] = -2