Ta có:
$\dfrac{a}{2x+3}+\dfrac{b}{x-1}$
$=\dfrac{(a+2b)x+3b-a}{(2x+3)(x-1)}$
Đồng nhất với biểu thức $\dfrac{1}{(2x+3)(x-1)}$, ta được:
$\left\{ \begin{array}{l}a+2b=0\\3b-a=1\end{array} \right.$
$↔ \left\{ \begin{array}{l}a=-\dfrac{2}{5}\\b=\dfrac{1}{5}\end{array} \right.$
$→ \int\limits^3_2 {\dfrac{1}{(2x+3)(x-1)}} \, dx$
$=\int\limits^3_2 {\dfrac{1}{5(x-1)}} \, dx-\int\limits^3_2 {\dfrac{2}{5(2x+3)}} \, dx$
$=\dfrac{1}{5}.ln|x-1||_2^3-\dfrac{1}{5}.ln|2x+3||_2^3$
$=\dfrac{1}{5}.ln\Bigg|\dfrac{x-1}{2x+3}\Bigg|\Bigg|_2^3$
$=\dfrac{1}{5}ln\dfrac{2}{9}-\dfrac{1}{5}ln\dfrac{1}{7}$
$=\dfrac{1}{5}ln\dfrac{14}{9}$
