Họ nguyên hàm của hàm số \(\int {\dfrac{{2x + 3}}{{2{x^2} - x - 1}}dx} \) là:

\(\frac{{2x + 3}}{{2{x^2} - x - 1}} = \frac{{2x + 3}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}}\)

Do đó, ta cần biến đổi \(\frac{{2x + 3}}{{2{x^2} - x - 1}} = \frac{a}{{2x + 1}} + \frac{b}{{x - 1}}\) để tính được nguyên hàm.

Ta có:

\(\begin{array}{l}\frac{a}{{2x + 1}} + \frac{b}{{x - 1}} = \frac{{a\left( {x - 1} \right) + b\left( {2x + 1} \right)}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}}\\ = \frac{{ax - a + 2bx + b}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}} = \frac{{\left( {a + 2b} \right)x - a + b}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}}\end{array}\)

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

 Do đó:

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