Answer
$\displaystyle \dfrac{1}{(2x+3)(4x-1)}=\dfrac{-\frac{1}{7}}{2x+3}+\dfrac{\frac{2}{7}}{4x-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: $$\displaystyle \frac{1}{(2x+3)(4x-1)}=\frac{A}{2x+3}+\frac{B}{4x-1} …(1)$$
Now, write the RHS with a common denominator.
$$\displaystyle \frac{1}{(2x+3)(4x-1)}=\dfrac{A(4x-1)+B(2x+3)}{(2x+3)(4x-1)}\\ \displaystyle \frac{1}{(2x+3)(4x-1)}=\dfrac{4Ax-A+2Bx+3B}{(2x+3)(4x-1)} \\ \displaystyle \frac{1}{(2x+3)(4x-1)}=\dfrac{(4A+2B)x+(-A+3B)}{(2x+3)(4x-1)}$$
Next, equate the numerators to obtain:
$$(4A+2B)x+(-A+3B)=1$$
Equate the like coefficients of the polynomials on the LHS and RHS to obtain:
$$4A+2B=0 ~~~(2) \\ -A+3B=1 ~~~(3)$$
Equation (2) yields: $$4A=-2B\\ 2B=-4A \\ B=\dfrac{-4A}{2}=-2A ~~~(4)$$
Substitute $B=-2A$ into equation (3) to solve for $A$. $$-A+3(-2A) =1\\ A=\dfrac{-1}{7}$$
Finally, substitute $A=\dfrac{-1}{7}$ into equation (4) to solve for $B$. $$B=-2A \\ B=-2 \times \dfrac{-1}{7} \\ B=\dfrac{2}{7}$$
Therefore, the equation (1) becomes:
$\displaystyle \dfrac{1}{(2x+3)(4x-1)}=\dfrac{-\frac{1}{7}}{2x+3}+\dfrac{\frac{2}{7}}{4x-1} $
