Nếu \(\int\limits_2^3 {\dfrac{{x + 2}}{{2{x^2} - 3x + 1}}{\rm{d}}x = a\ln 5 + b\ln 3 + 3\ln 2} \)\(\left( {a,b \in \mathbb{Q}} \right)\) thì giá trị của \(P = 2a - b\) là

Ta có \(\dfrac{{x + 2}}{{2{x^2} - 3x + 1}} = \dfrac{{x + 2}}{{\left( {x - 1} \right)\left( {2x - 1} \right)}} = \dfrac{A}{{x - 1}} + \dfrac{B}{{2x - 1}}\).

\(\begin{array}{l} \Leftrightarrow \dfrac{{x + 2}}{{2{x^2} - 3x + 1}} = \dfrac{{A\left( {2x - 1} \right) + B\left( {x - 1} \right)}}{{2{x^2} - 3x + 1}}\\ \Leftrightarrow x + 2 = \left( {2A + B} \right)x - A - B\end{array}\)

Đồng nhất hệ số ta có \(\left\{ \begin{array}{l}2A + B = 1\\ - A - B = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = 3\\B =  - 5\end{array} \right.\), khi đó ta có \(\dfrac{{x + 2}}{{2{x^2} - 3x + 1}} = \dfrac{3}{{x - 1}} - \dfrac{5}{{2x - 1}}\).

\(\begin{array}{l} \Rightarrow \int\limits_2^3 {\dfrac{{x + 2}}{{2{x^2} - 3x + 1}}{\rm{d}}x} \\ = \int\limits_2^3 {\left( {\dfrac{3}{{x - 1}} - \dfrac{5}{{2x - 1}}} \right)dx} \\ = \left. {\left( {3\ln \left| {x - 1} \right| - \dfrac{5}{2}\ln \left| {2x - 1} \right|} \right)} \right|_2^3\\ = 3\ln 2 - \dfrac{5}{2}\ln 5 - 3\ln 1 + \dfrac{5}{2}\ln 3\\ =  - \dfrac{5}{2}\ln 5 + \dfrac{5}{2}\ln 3 + 3\ln 2\end{array}\)

\( \Rightarrow a =  - \dfrac{5}{2},\,\,b = \dfrac{5}{2}\).

Vậy \(P = 2a - b = 2.\dfrac{{ - 5}}{2} - \dfrac{5}{2} =  - \dfrac{{15}}{2}\).