\(\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-3\right)=4\\ \Leftrightarrow x^3+1-x^3+3x=4\\ \Leftrightarrow3x=1-4=-3\\ \Leftrightarrow x=-3:3=-1\)

\(\Rightarrow x\in\left\{3;9\right\}\)

x = 1

\(\Rightarrow A=\dfrac{1-1}{1^2-9}+\dfrac{1}{1+3}=\dfrac{0}{-8}+\dfrac{1}{4}=\dfrac{1}{4}\)