If $\frac{9x-7}{\left(x+3\right)\left(x^2+1\right)}=\frac{A}{x+3}+\frac{Bx+C}{x^2+1}$, where $A, B, C\in ℝ,$ $A+B+C=$

Given, $\frac{9x-7}{\left(x+3\right)\left(x^2+1\right)}=\frac{A}{x+3}+\frac{Bx+C}{x^2+1}$

$\frac{9x-7}{\left(x+3\right)\left(x^2+1\right)}=\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x+3\right)}{\left(x+3\right)\left(x^2+1\right)}$

$9x-7=A\left(x^2+1\right)+\left(Bx+C\right)\left(x+3\right)$

$9x-7=A{x}^2+A+B{x}^2+Cx+3Bx+3C$

$9x-7=x^2\left(A+B\right)+x\left(3B+C\right)+A+3C$

Comparing on both sides

$A+B=0, 3B+C=9, A+3C=-7$

Solving these equations

$A=-\frac{17}{5}, B=\frac{17}{5}, C=-\frac{6}{5}$

Value of $A+B+C=-\frac{17}{5}+\frac{17}{5}-\frac{6}{5}$

Hence, $A+B+C=-\frac{6}{5}$