Tích phân \(\int\limits_{ - 1}^1 {\dfrac{x}{{{x^2} - 5\left| x \right| + 6}}dx} \) bằng:

Cách 1:

\(\int\limits_{ - 1}^1 {\dfrac{x}{{{x^2} - 5\left| x \right| + 6}}dx}  = \int\limits_{ - 1}^0 {\dfrac{x}{{{x^2} + 5x + 6}}dx}  + \int\limits_0^1 {\dfrac{x}{{{x^2} - 5x + 6}}dx}  = {I_1} + {I_2}\)

Xét tích phân \({I_1} = \int\limits_{ - 1}^0 {\dfrac{x}{{{x^2} + 5x + 6}}dx} \) ta có:

$\begin{array}{l}\dfrac{x}{{{x^2} + 5x + 6}} = \dfrac{x}{{\left( {x + 2} \right)\left( {x + 3} \right)}} = \dfrac{A}{{x + 2}} + \dfrac{B}{{x + 3}} = \dfrac{{Ax + 3A + Bx + 2B}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\ \Rightarrow \left\{ \begin{array}{l}A + B = 1\\3A + 2B = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A =  - 2\\B = 3\end{array} \right. \Rightarrow \dfrac{x}{{{x^2} + 5x + 6}} = \dfrac{{ - 2}}{{x + 2}} + \dfrac{3}{{x + 3}}\\ \Rightarrow {I_1} = \int\limits_{ - 1}^0 {\left( {\dfrac{{ - 2}}{{x + 2}} + \dfrac{3}{{x + 3}}} \right)dx}  = \left. {\left( { - 2\ln \left| {x + 2} \right| + 3\ln \left| {x + 3} \right|} \right)} \right|_{ - 1}^0\\ =  - 2\ln 2 + 3\ln 3 - 3\ln 2 = 3\ln 3 - 5\ln 2\end{array}$ 

Xét tích phân \({I_2} = \int\limits_0^1 {\dfrac{x}{{{x^2} - 5x + 6}}dx} \) ta có :

\(\begin{array}{l}\dfrac{x}{{{x^2} - 5x + 6}} = \dfrac{x}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{A}{{x - 2}} + \dfrac{B}{{x - 3}} = \dfrac{{Ax - 3A + Bx - 2B}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\ \Rightarrow \left\{ \begin{array}{l}A + B = 1\\ - 3A - 2B = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A =  - 2\\B = 3\end{array} \right. \Rightarrow \dfrac{x}{{{x^2} - 5x + 6}} = \dfrac{{ - 2}}{{x - 2}} + \dfrac{3}{{x - 3}}\\ \Rightarrow {I_2} = \int\limits_0^1 {\left( {\dfrac{{ - 2}}{{x - 2}} + \dfrac{3}{{x - 3}}} \right)dx}  = \left. {\left( { - 2\ln \left| {x - 2} \right| + 3\ln \left| {x - 3} \right|} \right)} \right|_0^1\\ = 3\ln 2 + 2\ln 2 - 3\ln 3 = 5\ln 2 - 3\ln 3\end{array}\)

\( \Rightarrow I = {I_1} + {I_2} = 0\)