The partial fraction for $\frac{px+q}{\left(x+a\right)\left(x^2+b^2\right)}$ is of type $\frac{A}{x+a}+\frac{Bx+C}{x^2+b^2}$

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

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

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

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

On comparing, the coefficients of $x^2, x$ and the constant term, we get

$A+B=0 ...\left(i\right)$

$3B+C=9 ...\left(ii\right)$ and

$A+3C=-7 ...\left(iii\right)$

On solving the above three equations, we get $A=-\frac{17}{5}, B=\frac{17}{5}, C=-\frac{6}{5}.$

Hence, $\frac{9x-7}{\left(x+3\right)\left(x^2+1\right)}=\frac{\left(-{\displaystyle \frac{17}{5}}\right)}{x+3}+\frac{\left({\displaystyle \frac{17}{5}}\right)x+\left(-{\displaystyle \frac{6}{5}}\right)}{x^2+1}$

$\Rightarrow \frac{9x-7}{\left(x+3\right)\left(x^2+1\right)}=\frac{-{\displaystyle 17}}{5\left(x+3\right)}+\frac{\left(17x-6\right)}{5\left(x^2+1\right)}.$