Biết rằng \(\int\limits_1^2 {\dfrac{{{x^3} - 1}}{{{x^2} + x}}dx = a + b\ln 3 + c\ln 2} \) với \(a,{\mkern 1mu} {\mkern 1mu} b,{\mkern 1mu} {\mkern 1mu} c\) là các số hữu tỉ. Tính \...

* Hướng dẫn giải

Ta có:

\(\begin{array}{*{20}{l}}{\int\limits_1^2 {\dfrac{{{x^3} - 1}}{{{x^2} + x}}dx} {\rm{\;}} = \int\limits_1^2 {\left( {x - 1 + \dfrac{{x - 1}}{{{x^2} + x}}} \right)dx} }\\{ = \int\limits_1^2 {\left( {x - 1} \right)dx} {\rm{\;}} + \int\limits_1^2 {\dfrac{{x - 1}}{{x\left( {x + 1} \right)}}dx} }\\{ = \dfrac{1}{2} + I}\end{array}\)

Giả sử \(\dfrac{{x - 1}}{{x\left( {x + 1} \right)}} = \dfrac{B}{x} + \dfrac{C}{{x + 1}}\)

\(\begin{array}{*{20}{l}}{ \Leftrightarrow \dfrac{{x - 1}}{{x\left( {x + 1} \right)}} = \dfrac{{B\left( {x + 1} \right) + Cx}}{{x\left( {x + 1} \right)}}}\\{ \Leftrightarrow \dfrac{{x - 1}}{{x\left( {x + 1} \right)}} = \dfrac{{\left( {B + C} \right)x + B}}{{x\left( {x + 1} \right)}}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{B + C = 1}\\{B = {\rm{\;}} - 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{B = {\rm{\;}} - 1}\\{C = 2}\end{array}} \right.}\end{array}\)

Khi đó ta có

\(\begin{array}{*{20}{l}}{I = \int\limits_1^2 {\dfrac{{x - 1}}{{x\left( {x + 1} \right)}}dx} {\rm{\;}} = \int\limits_1^2 {\dfrac{{ - 1}}{x}dx} {\rm{\;}} + \int\limits_1^2 {\dfrac{2}{{x + 1}}dx} }\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = \left. { - \ln \left| x \right|} \right|_1^2 + \left. {2\ln \left| {x + 1} \right|} \right|_1^2}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = {\rm{\;}} - \ln 2 + 2\ln 3 - 2\ln 2}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = 2\ln 3 - 3\ln 2}\end{array}\)

\( \Rightarrow \int\limits_1^2 {\dfrac{{{x^3} - 1}}{{{x^2} + x}}dx} {\rm{\;}} = \dfrac{1}{2} + 2\ln 3 - 3\ln 2\) \( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = \dfrac{1}{2}}\\{b = 2}\\{c = {\rm{\;}} - 3}\end{array}} \right.\).

Vậy \(2a + 3b - 4c = 2.\dfrac{1}{2} + 3.2 - 4.\left( { - 3} \right) = 19\).

Chọn D.