Chapter 5, Systems of Equations and Inequalities - Section 5.3 - Partial Fractions - 5.3 Exercises - Page 462: 29

$\frac{2x}{(2x+3)^2}=\frac{1}{2x+3}+\frac{-3}{(2x+3)^2}$,

$\frac{2x}{4x^2+12x+9}=$ factorising $4x^2+12x+9=(2x+3)(2x+3)=(2x+3)^2$, $\frac{2x}{(2x+3)^2}=\frac{A}{2x+3}+\frac{B}{(2x+3)^2}$, $A(2x+3)+B=2x$, $2A=2, A=1$, $3A+B=0, B=-3$. $\frac{2x}{(2x+3)^2}=\frac{1}{2x+3}+\frac{-3}{(2x+3)^2}$,