Biết \(\int\limits_0^1 {\dfrac{{3x - 1}}{{{x^2} + 6x + 9}}dx}  = 3\ln \dfrac{a}{b} - \dfrac{5}{6}\), trong đó \(a,\,\,b\) là các số nguyên dương và \(\dfrac{a}{b}\) tối giản. Khi đó \({a^2} - {b^2}\) bằng:

Ta có:

\(\begin{array}{l}\,\,\,\,\,\dfrac{{3x - 1}}{{{x^2} + 6x + 9}} = \dfrac{{3x - 1}}{{{{\left( {x + 3} \right)}^2}}} = \dfrac{A}{{{{\left( {x + 3} \right)}^2}}} + \dfrac{B}{{x + 3}}\\ \Leftrightarrow \dfrac{{3x - 1}}{{{x^2} + 6x + 9}} = \dfrac{{A + B\left( {x + 3} \right)}}{{{{\left( {x + 3} \right)}^2}}}\\ \Leftrightarrow \dfrac{{3x - 1}}{{{x^2} + 6x + 9}} = \dfrac{{Bx + A + 3B}}{{{{\left( {x + 3} \right)}^2}}}\end{array}\)

Đồng nhất hệ số ta có \(\left\{ \begin{array}{l}B = 3\\A + 3B =  - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}B = 3\\A =  - 10\end{array} \right.\).

\(\begin{array}{l} \Rightarrow \dfrac{{3x - 1}}{{{x^2} + 6x + 9}} =  - \dfrac{{10}}{{{{\left( {x + 3} \right)}^2}}} + \dfrac{3}{{x + 3}}\\ \Rightarrow \int\limits_0^1 {\dfrac{{3x - 1}}{{{x^2} + 6x + 9}}dx}  =  - 10\int\limits_0^1 {\dfrac{{dx}}{{{{\left( {x + 3} \right)}^2}}}}  + 3\int\limits_0^1 {\dfrac{{dx}}{{x + 3}}} \\ \Leftrightarrow \int\limits_0^1 {\dfrac{{3x - 1}}{{{x^2} + 6x + 9}}dx}  = 10.\left. {\dfrac{1}{{x + 3}}} \right|_0^1 + 3\left. {\ln \left| {x + 3} \right|} \right|_0^1\\ \Leftrightarrow \int\limits_0^1 {\dfrac{{3x - 1}}{{{x^2} + 6x + 9}}dx}  = 10\left( {\dfrac{1}{4} - \dfrac{1}{3}} \right) + 3\left( {\ln 4 - \ln 3} \right)\\ \Leftrightarrow \int\limits_0^1 {\dfrac{{3x - 1}}{{{x^2} + 6x + 9}}dx}  = 3\ln \dfrac{4}{3} - \dfrac{5}{6}\end{array}\)

Khi đó ta có \(a = 4,\,\,b = 3\).

Vậy \({a^2} - {b^2} = {4^2} - {3^2} = 7\).