$\because \frac{d}{d x}\left(\frac{x^2}{(x+2)(2 x+3)}\right)=\frac{\mathrm{A}}{(x+2)^2}+\frac{\mathrm{B}}{(2 x+3)^2}$
$\Rightarrow\left(\frac{7 x^2+12 x}{(x+2)(2 x+3)}\right)=\frac{\mathrm{A}}{(x+2)^2}+\frac{\mathrm{B}}{(2 x+3)^2}$ ...(i)
The form of the partial fraction decomposition is
$\frac{7 x^2+12 x}{(x+2)^2(2 x+3)^2}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B}}{(x+2)^2}+\frac{\mathrm{C}}{(2 x+3)}+\frac{\mathrm{D}}{(2 x+3)^2}$ ...(ii)
$\begin{array}{r}\Rightarrow 7 x^2+12 x=\mathrm{A}(x+2)(2 x+3)^2+\mathrm{B}(2 x+3)^2+ \\ \mathrm{C}(x+2)^2(2 x+3)+\mathrm{D}(x+2)^2\end{array}$
$\begin{aligned} & \Rightarrow 7 x^2+12 x=x^3(4 \mathrm{~A}+2 \mathrm{C})+x^2(20 \mathrm{~A}+4 \mathrm{~B}+11 \mathrm{C}+\mathrm{D})+ \\ & x(33 \mathrm{~A}+12 \mathrm{~B}+20 \mathrm{C}+4 \mathrm{D})+18 \mathrm{~A}+9 \mathrm{~B}+12 \mathrm{C}+4 \mathrm{D}\end{aligned}$
Comparing the coefficients, we get:
$\begin{aligned} & 4 A+2 C=0 \\ & 20 A+4 B+11 C+D=7\end{aligned}$
$33 \mathrm{~A}+12 \mathrm{~B}+20 \mathrm{C}+4 \mathrm{D}=12$ ...(iii)
$18 \mathrm{~A}+9 \mathrm{~B}+12 \mathrm{C}+4 \mathrm{D}=0$
Solving above system of equation, we get $A=0, B=4, C=0, D=-9$
$\begin{aligned} & \therefore \frac{7 x^2+12 x}{(x+2)^2(2 x+3)^2}=\frac{4}{(x+2)^2}-\frac{9}{(2 x+3)^2} \\ & \therefore \frac{d}{d x}\left(\frac{x^2}{(x+2)(2 x+3)}\right)=\frac{4}{(x+2)^2}+\frac{(-9)}{(2 x+3)^2}\end{aligned}$
From $\mathrm{eq}^{\mathrm{n}}$ (i)
$A=4, B=-9 \Rightarrow A+B=4-9=-5$