Verified answer

When you are dealing with partial fraction decomposition with a repeated factor, in this case, you want something of the form

A / (2x+3) + B / (2x + 3)^2

Where A and B are numbers. You may ask: why not an Ax+B in the numerator on the second fraction, but you already have that, that's your original problem. So in this case, given the numerators, we have

A(2x+3) + B(1) = 2x

2Ax + 3A + B = 2x

2Ax = 2x

3A+B = 0

2A = 1

3A + B = 0

A = 1/2

3(1/2) + B = 0

B = -3/2

So your partial fraction decomposition becomes

(3/2) / (2x + 3) - (3/2) / (2x+3)^2

4x^2 + 12x + 9 = (2x+3)^2

so the partial fraction would look like this:

a/(2x+3) + b/(2x+3)^2 because the denominator is a perfect square.

= [a(2x+3) + b]/(2x+3)^2 so now you need only consider the numerator.

2ax + 3a + b = 2x -->

2a = 1 and 3a + b = 0 -->

a = 1/2 and thus b = -3/2

The partial fraction becomes

(1/2)/[2x+3] - (3/2)/[2x+3]^2