$A= 8 - 8 +\dfrac{8x+6}{x^2+1}=8+\dfrac{-8x^2+8x-2}{x^2+1}=8-\dfrac{2(2x+1)^2}{x^2+1} \le 8$$ A=-2 +2+ \dfrac{8x+6}{x^2+1}=-2+\dfrac{2x^2+8x+8}{x^2 +1}=-2+\dfrac{2(x+2)^2}{x^2+1} \ge -2$
$A= 8 - 8 +\dfrac{8x+6}{x^2+1}=8+\dfrac{-8x^2+8x-2}{x^2+1}=8-\dfrac{2(2x+1)^2}{x^2+1} \le 8$$A=-2 +2\dfrac{8x+6}{x^2+1}=-2+\dfrac{2x^2+8x+8}{x^2 +1}=-2+\dfrac{2(x^2+2)^2}{x^2+1} \ge -2$
$A= 8 - 8 +\dfrac{8x+6}{x^2+1}=8+\dfrac{-8x^2+8x-2}{x^2+1}=8-\dfrac{2(2x+1)^2}{x^2+1} \le 8$$
A=-2 +2
+ \dfrac{8x+6}{x^2+1}=-2+\dfrac{2x^2+8x+8}{x^2 +1}=-2+\dfrac{2(x+2)^2}{x^2+1} \ge -2$