A value of $lpha$ such that

$\int_{\alpha}^{\alpha+1} \frac{d x}{(x+\alpha)(x+\alpha+1)}$

$=\int_{\alpha}^{\alpha+1}\left[\frac{1}{x+\alpha}-\frac{1}{x+\alpha+1}\right] d x$ [Using partial fraction]

$=\left.\log \left(\frac{(x+\alpha)}{(x+\alpha+1)}\right)\right|_{\alpha} ^{\alpha+1}=\log \left(\frac{2 \alpha+1}{2 \alpha+2} \cdot \frac{2 \alpha+1}{2 \alpha}\right)$

$=\log \frac{9}{8}$ (Given)

So, $\frac{(2 \alpha+1)^{2}}{\alpha(\alpha+1)}=\frac{9}{2} \Rightarrow 8 \alpha^{2}+8 \alpha+2=9 \alpha^{2}+9 \alpha$

$\Rightarrow \alpha^{2}+\alpha-2=0 \Rightarrow \alpha=1,-2$