Question

Find the partial fraction
$\frac{2x+1}{(3x+2)(4x^2+5x+6)}$.

Answer 1

Let $\frac{2x+1}{(3x+2)(4x^2+5x+6)}=\frac{A}{(3x+2)}\frac{Bx+C}{(4x^2+5x+6)}$
then $2x+1=A(4x^2+5x+6)+(Bx+C)(3x+2)$
where $A,B,C$ are constants.
For $A$, let $3x+2=0$,
i.e., $x=-2/3$
$2(-\frac{2}{3})+1=A\{4.\frac{4}{9}-\frac{10}{3}+6\}+\{B(-\frac{2}{3})+C\}\left(0\right)$
$-\frac{1}{3}=A\left(\frac{40}{9}\right)\Rightarrow A=-\frac{3}{40}$