Điều kiện: $|x|<2,x\ne0,x\ne\pm\sqrt3$
Đặt: $t=\log_{x^2+1}(4-x^2) \Rightarrow (4-x^2)=(x^2+1)^t$
$\Rightarrow (x^2+1)=(4-x^2)^{\frac{1}{t}} \Rightarrow \log_{4-x^2}(x^2+1)=\dfrac{1}{t}$
Khi đó:
$y=\left|t+\dfrac{1}{t}\right|=\dfrac{|t^2+1|}{|t|}\ge\dfrac{2|t|}{|t|}=2$
$\min y=2 \Leftrightarrow \left[\begin{array}{l}t=1\\t=-1\end{array}\right.\Leftrightarrow \left[\begin{array}{l}x^2+1=4-x^2\\x^2+1=\dfrac{1}{4-x^2}\end{array}\right.\Leftrightarrow \left[\begin{array}{l}x=\pm\sqrt{\dfrac{3}{2}}\\x=\pm\sqrt{\dfrac{3+\sqrt{21}}{2}}\end{array}\right.$