Nguyên hàm của $\Large \int \dfrac{11-x}{(2x-1)(3x+2)}dx$ là $\Large \

Nguyên hàm của $\Large \int \dfrac{11-x}{(2x-1)(3x+2)}dx$ là

• A. $\Large \dfrac{3}{2}ln\left | 2x-1 \right |-\dfrac{5}{3}ln\left | 3x+2 \right |+C$
• B. $\Large 3ln\left | 2x-1 \right |-5ln\left | 3x+2 \right |+C$
• C. $\Large \dfrac{3}{2}ln\left | 2x-1 \right |+\dfrac{5}{3}ln\left | 3x+2 \right |+C$
• D. $\Large 3ln\left | 2x-1 \right |+5ln\left | 3x+2 \right |+C$

Ta có: $\Large \dfrac{11-x}{(2x-1)(3x+2)}=\dfrac{A}{2x-1}+\dfrac{B}{3x+2}$

Quy đồng vế phải $\Large \dfrac{A}{2x-1}+\dfrac{B}{3x+2}=\dfrac{3Ax+2A+2Bx-B}{(2x-1)(3x+2)}=\dfrac{(3A+2B)x+2A-B}{(2x-1)(3x+2)}$

Đồng nhất tử thức, tức là cho $\Large \left\{\begin{matrix}
3A+2B=-1\\ 
2A-B=11
\end{matrix}\right.$ ta được $\Large \left\{\begin{matrix}
A=3\\ 
B=-5
\end{matrix}\right.$

Suy ra: $\Large \int \dfrac{11-x}{(2x-1)(3x+2)}dx=\int\left ( \dfrac{3}{2x-1}+\dfrac{-5}{3x+2} \right )dx$

$\Large =\int \dfrac{3}{2}.\dfrac{d(2x-1)}{2x-1}-\int \dfrac{5}{3}.\dfrac{d(3x+2)}{3x+2}$

$\Large =\dfrac{3}{2}ln\left | 2x-1 \right |-\dfrac{5}{3}ln\left | 3x+2 \right |+C$