The partial fractions of $\frac{x^2+13x+15}{\left(2x+3\right){\left(x+3\right)}^2}$  are

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

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

$=A\left[x^2+6x+9\right]+B\left[2x^2+9x+9\right]+2Cx+3C$

$=\left(A+2B\right)x^2+\left(6A+9B+2C\right)x+9A+9B+3C$

$A+2B=1\cdots \cdots \cdots \left(1\right)$

$6A+9B+2C=13\cdots \cdots \cdots \left(2\right)$

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

From equations $\left(3\right)\&\left(2\right)$

$2\left(9A+9B+3C=15\right)$

$\underset{\_}{3\left(6A+9B+2C=13\right)}$

$-9B=-9$

$B=1$

$\left(1\right)\Rightarrow A+2\left(1\right)=1\Rightarrow A=1-2=-1$

$\left(2\right)\Rightarrow 6\left(-1\right)+9\left(1\right)+2C=13$

$3+2C=13$

$2C=10$

$C=5$

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