$\frac{x^2+13x+15}{(2x+3){(x+3)}^2}=$

$\frac{x^2+13x+15}{(2x+3){(x+3)}^2}=\frac{A}{2x+3}+\frac{B}{x+3}+\frac{C}{{(x+3)}^2}$

$x^2+13x+15=A{(x+3)}^2+B(x+3)(2x+3)+C(2x+3)$

$\text{if }x=\frac{-3}{2}$

$\frac{9}{4}+13\left(\frac{-3}{2}\right)+15=A{(\frac{-3}{2}+3)}^2$

$A=-1$

$\text{if }x=-3$

$9+13(-3)+15=c(2(-3)+3)$

$C=5$

$\text{if }x=0$

$15=A\left(9\right)+B\left(3\right)\left(3\right)+C\left(3\right)$

$15=9A+9B+3C$

$15=-9+9B+15\Rightarrow$

$B=1$

$\therefore \frac{x^2+13x+15}{(2x+3){(x+3)}^2}=\frac{-1}{2x+3}+\frac{1}{x+3}+\frac{5}{{(x+3)}^2}$