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

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

By comparing  $x^2$ coefficient , $A+B=0$

Put  $x=-3$then  $A=-\frac{17}{5}\Rightarrow B=\frac{17}{5}$

Compare coefficient of  $x$on both the side  $3B+C=9$.

$\Rightarrow C=-3B+9$

$=9-3\left(\frac{17}{5}\right)$

$=-\frac{6}{5}$

$\therefore A+B+C=-\frac{6}{5}$