If $\frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{A}{x^2+3}+\frac{B}{x^2+2}$  then $\left(A,B\right)=$

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

$x^2+1=A\left(x^2+2\right)+B\left(x^2+3\right)$

$x^2+1=\left(A+B\right)x^2+2A+3B$

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

$2A+3B=1\cdots \cdots \left(2\right)$

Equations $\left(1\right)\&\left(2\right)$

$2A+3B=1$

$\underset{\_}{2\left(A+B=1\right)}$

$B=-1$

Equation $\left(1\right)\Rightarrow A=1-B$

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

$\therefore \left(A,B\right)=\left(2,-1\right)$