Phương pháp giải:

\(\frac{\left( 3x-1 \right)}{{{x}^{2}}+6x+9}=\frac{3x-1}{{{\left( x+3 \right)}^{2}}}=\frac{A}{{{\left( x+3 \right)}^{2}}}+\frac{B}{x+3}\), đồng nhất hệ số tìm A, B.

Lời giải chi tiết:

\(I=\int\limits_{0}^{1}{\dfrac{\left( 3x-1 \right)dx}{{{x}^{2}}+6x+9}}=\int\limits_{0}^{1}{\dfrac{3x-1}{{{\left( x+3 \right)}^{2}}}dx}\)

Ta có:

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

Chọn C.