Given:

$\frac{x^2+3}{\left(x^2+1\right)\left(x^2+2\right)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}$

$\Rightarrow \frac{x^2+3}{\left(x^2+1\right)\left(x^2+2\right)}=\frac{\left(Ax+B\right)\left(x^2+2\right)+\left(Cx+D\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2+2\right)}$

$\Rightarrow x^2+3=\left(Ax+B\right)\left(x^2+2\right)+\left(Cx+D\right)\left(x^2+1\right)$

$\Rightarrow x^2+3=\left(A+C\right)x^3+\left(B+D\right)x^2+\left(2A+C\right)x+\left(2B+D\right)$

On comparing coefficients both sides, we get

$$\begin{gathered} \left\{\begin{array}{l}A+C=0 ....\left(\mathrm{i}\right) \\ B+D=1 ....\left(\mathrm{ii}\right) \\ 2A+C=0 ....\left(\mathrm{iii}\right)\\ 2B+D=3 ....\left(\mathrm{iv}\right)\end{array}\right. \end{gathered}$$

On solving, we get

$A=0, B=2, C=0, D=-1$

So,

$A+B+C+D=1$