If $\frac{{(x+2)}^2}{x^3+2x^2+3x}=\frac{A}{x}+\frac{Bx+C}{x^2+2x+3},$then

${(x+2)}^2=A(x^2+2x+3x)+(Bx+1)\left(x\right)$

$putx=04=3A+0$

$A=\frac{4}{3}$

Coefficient of $x^2$

$\begin{array}{l}1=A+B\\ 1=\frac{4}{3}+B\\ -\frac{1}{3}=B\end{array}$

Comparing coefficient of ‘x’

$\begin{array}{l}4=2A+C\\ 4=2\left(\frac{4}{3}\right)+C\\ \frac{4}{3}=C\end{array}$