Lời giải của Tự Học 365

Ta có:

$\begin{array}{l}\dfrac{{2x + 3}}{{2{x^2} - x - 1}} = \dfrac{{2x + 3}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}} = \dfrac{a}{{2x + 1}} + \dfrac{b}{{x - 1}} \\ \Leftrightarrow \dfrac{{2x + 3}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}} = \dfrac{{a\left( {x - 1} \right) + b\left( {2x + 1} \right)}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}}\\ \Leftrightarrow \dfrac{{2x + 3}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}} = \dfrac{{\left( {a + 2b} \right)x - a + b}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}} \Leftrightarrow \left\{ \begin{array}{l}2 = a + 2b\\3 =  - a + b\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a =  - \dfrac{4}{3}\\b = \dfrac{5}{3}\end{array} \right.\end{array}$

 Do đó:

$\begin{array}{l}\int {\dfrac{{2x + 3}}{{2{x^2} - x - 1}}dx}  = \int {\left[ { - \dfrac{4}{3}.\dfrac{1}{{\left( {2x + 1} \right)}} + \dfrac{5}{3}.\dfrac{1}{{\left( {x - 1} \right)}}} \right]dx}  \\ =  - \dfrac{4}{3}\int {\dfrac{1}{{\left( {2x + 1} \right)}}dx}  + \dfrac{5}{3}\int {\dfrac{1}{{\left( {x - 1} \right)}}dx} \\ =  - \dfrac{4}{3}.\dfrac{1}{2}\ln \left| {2x + 1} \right| + \dfrac{5}{3}\ln \left| {x - 1} \right| + C =  - \dfrac{2}{3}\ln \left| {2x + 1} \right| + \dfrac{5}{3}\ln \left| {x - 1} \right| + C\end{array}$

Đáp án cần chọn là: b