From question, the equation given is:

\(\mathop \smallint \nolimits_\alpha ^{\alpha + 1} \frac{{{\rm{d}}x}}{{\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}} = {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left( {\frac{9}{8}} \right)\)

Let’s assume the integral as ‘I’,

\(\Rightarrow I = \mathop \smallint \nolimits_\alpha ^{\alpha + 1} \frac{{{\rm{d}}x}}{{\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}\)

On putting, t = (x + α),

\(\Rightarrow \frac{{dt}}{{dx}} = 1 \Rightarrow dt = dx\)

For limits:

⇒ x = α + 1 ⇒ t = 2α + 1

⇒ x = α ⇒ t = 2α

On substituting the integral becomes,

\(\Rightarrow I = \mathop \smallint \nolimits_{2\alpha }^{2\alpha + 1} \frac{{{\rm{d}}t}}{{\left( t \right)\left( {t + 1} \right)}}\)

On breaking the above integral,

\(\Rightarrow I = \mathop \smallint \nolimits_{2\alpha }^{2\alpha + 1} \left( {\frac{1}{t} - \frac{1}{{t + 1}}} \right)dt\)

\(\left[ {\smallint \frac{1}{x}dx = {\rm{ln}}\left| x \right| + C} \right]\)

\(\Rightarrow I=\left[ \ln \left| t \right|-\ln \left| t+1 \right| \right]_{2\alpha }^{2\alpha +1}\)

\(\because \left[ \text{ln}\left( \frac{x}{y} \right)=\text{ln}\left( x \right)-\text{ln}\left( y \right) \right]\)

\(\Rightarrow I = \left[ {\ln \left( {\frac{t}{{t + 1}}} \right)} \right]_{2\alpha }^{2\alpha + 1}\)

On direct substitution,

\(\Rightarrow I = \left[ {\ln \left( {\frac{{2\alpha + 1}}{{2\alpha + 1 + 1}}} \right)} \right] - \left[ {\ln \left( {\frac{{2\alpha }}{{2\alpha + 1}}} \right)} \right]\)

From question,

\(\Rightarrow \left[ {\ln \left( {\frac{{2\alpha + 1}}{{2\alpha + 1 + 1}}} \right)} \right] - \left[ {\ln \left( {\frac{{2\alpha }}{{2\alpha + 1}}} \right)} \right] = {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left( {\frac{9}{8}} \right)\)

\(\Rightarrow \left[ {\ln \left( {\frac{{2\alpha + 1}}{{2\alpha + 2}}} \right)} \right] - \left[ {\ln \left( {\frac{{2\alpha }}{{2\alpha + 1}}} \right)} \right] = {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left( {\frac{9}{8}} \right)\)

\(\because \left[ \text{ln}\left( \frac{x}{y} \right)=\text{ln}\left( x \right)-\text{ln}\left( y \right) \right]\)

\(\Rightarrow \ln \left( \frac{\left( \frac{2\alpha +1}{2\alpha +2} \right)}{\left( \frac{2\alpha }{2\alpha +1} \right)} \right)=\text{lo}{{\text{g}}_{\text{e}}}\left( \frac{9}{8} \right)\)

\(\Rightarrow \ln \left( \frac{2\alpha +1}{2\alpha +2}\times \frac{2\alpha +1}{2\alpha } \right)=\text{lo}{{\text{g}}_{\text{e}}}\left( \frac{9}{8} \right)\)

\(\Rightarrow \ln \left( \frac{{{\left( 2\alpha +1 \right)}^{2}}}{2\alpha \left( 2\alpha +2 \right)} \right)=\text{lo}{{\text{g}}_{\text{e}}}\left( \frac{9}{8} \right)\)

On cancelling ln and loge,

\(\Rightarrow \frac{{{\left( 2\alpha +1 \right)}^{2}}}{2\alpha \left( 2\alpha +2 \right)}=\frac{9}{8}\)

∵ [(a + b)2 = a2 + b2 + 2ab]

\(\Rightarrow \frac{4{{\alpha }^{2}}+1+4\alpha }{4{{\alpha }^{2}}+4\alpha }=\frac{9}{8}\)

⇒ 8(4α2 + 1 + 4α) = 9(4α2 + 4α)

⇒ 32α2 + 8 + 32α = 36α2 + 36α

⇒ 36α2 + 36α – 32α2 – 8 – 32α = 0

⇒ 4α2 + 4α – 8 = 0

⇒ 4 (α2 + α – 2) = 0

⇒ α2 + α – 2 = 0

⇒ α2 – α + 2α – 2 = 0

⇒ α(α – 1) + 2(α – 1) = 0

⇒ (α + 2)(α – 1) = 0

∴ α = -2, 1