We have to express 4x/ (4x^2 – 9) as a sum of fractions. We see that 4x^2 – 9 = (2x – 3) (2x + 3)\

Let A / (2x – 3) + B/ (2x + 3) = 4x/ (4x^2 – 9)

=> [(2x + 3) A + B (2x – 3)]/ (4x^2 – 9) = 4x/ (4x^2 – 9)

=> 2xA + 3A + 2xB – 3B = 4x

Equate terms with x and numeric terms

=> 2A + 2B = 4

=> A + B = 2

3A – 3B = 0

=> A – B = 0

Adding A + B = 2 and A – B = 0

=> 2A = 2

=> A = 1

B = 2 – A = 1

**Therefore 4x/ (4x^2 – 9) = 1 / (2x – 3) + 1 / (2x + 3)**

To Express 4x/ (4x^2 – 9) as a sum of fractions.

We know the denominator 4x^2-9 = (2x)^2-3^2 = (2x+3)(2x-3).

Let 4x/ (4x^2 – 9) = A/(2x-3) +B/(2x-3).

We multiply both sides by ((2x-3)(2x+3):

4x = A((2x+3)+B(2x-3).

4x = 2x(A+B) 3(A_B). If this is an identity then can equate the coefficients of like terms on both sides.

=> 2(A+B) = 4. Or A+B = 4/2 = 2.

A-B = 0, Or A - B. SO A+B = 2A =2 , Or A = 1. So B = 1.

Therefore 4x/(4x^2-9) = 1/{2(2x+3)}+1/{2(2x-3)}.