Cho \(\int {\frac{{2x - 13}}{{{x^2} - x - 2}}} \;{\rm{d}}x = a\ln \left| {x + 1} \right| + b\ln \left| {x - 2} \right| + C\) với \(a,\,\,b \in \mathbb{Q}.\)

Ta có \(\int {\frac{{2x - 13}}{{{x^2} - x - 2}}} \;{\rm{d}}x = \int {\frac{{2x - 13}}{{\left( {x + 1} \right)\left( {x - 2} \right)}}} {\rm{d}}x = \int {\left( {\frac{5}{{x + 1}} - \frac{3}{{x - 2}}} \right)} {\rm{d}}x\)

\( = 5.\int {\frac{1}{{x + 1}}} \;{\rm{d}}x - 3 \cdot \int {\frac{1}{{x - 1}}} \;{\rm{d}}x = 5\ln \left| {x + 1} \right| - 3\ln \left| {x - 2} \right| + C.\)

Vậy \(a = 5\,;\,\,b = - 3 \Rightarrow a - b = 8.\)

Chọn D.