Answer
$\dfrac{x}{(3x-2)(2x+1)}=\dfrac{\frac{2}{7}}{3x-2}+\dfrac{\frac{1}{7}}{2x+1}$
Work Step by Step
We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has non-repeated irreducible quadratic factor. So, we will write the given rational expression into partial fraction form as: $$\displaystyle \frac{x}{(3x-2)(2x+1)}=\frac{A}{3x-2}+\frac{B}{2x+1} …(1)$$
Now, write the RHS with a common denominator.
$$\displaystyle \frac{x}{(3x-2)(2x+1)}=\dfrac{A(2x+1)+B(3x-2)}{(3x-2)(2x+1)} \\ \displaystyle \frac{x}{(3x-2)(2x+1)}=\dfrac{2Ax+A+3Bx-2B}{(3x-2)(2x+1)}\\ \displaystyle \frac{x}{(3x-2)(2x+1)}=\dfrac{(2A+3B)x+(A-2B)}{(3x-2)(2x+1)} $$
Next, equate the numerators to obtain:
$$(2A+3B)x+(A-2B)=x$$
Equate the like coefficients of the polynomials on the LHS and RHS to obtain: $$2A+3B=1 ~~~(2) \\ A-2B=0 ~~~(3)$$
Equation (3) yields: $$-2B=-A \\ B=\dfrac{A}{2} ~~~(4) $$
Substitute $B=\dfrac{A}{2}$ into equation (2) to solve for $A$. $$2A+3(\dfrac{A}{2}) =1 \\ 2A+\dfrac{3A}{2}=1 \\ \dfrac{4A+3A}{2}=1\\ \dfrac{7}{2} A=1\\ A=\dfrac{2}{7}$$
Finally, substitute $A=\dfrac{2}{7}$ into equation (4) to solve for $B$. $$B=\dfrac{A}{2} \\ B=\dfrac{2}{14}=\dfrac{1}{7} $$
Therefore, the equation (1) becomes: $\displaystyle \dfrac{x}{(3x-2)(2x+1)}=\dfrac{\frac{2}{7}}{3x-2}+\dfrac{\frac{1}{7}}{2x+1}$
