To solve the inequality:

$\frac{|x^2 + 1|}{|x^2 - 5|} \leq 2$

Step-by-step solution:

1. Break down the inequality into two cases based on the absolute values.

   Since $x^2 + 1$ and $x^2 - 5$ are quadratic expressions, they are always non-negative. This is because:

   • $x^2 + 1 \geq 1$ for all real $x$ (since $x^2 \geq 0$),
   • $x^2 - 5 \geq 0$ only for $x^2 \geq 5$ (i.e., $x \geq \sqrt{5}$ or $x \leq -\sqrt{5}$).

   Therefore, the absolute values can be replaced by the expressions themselves depending on the sign of $x^2 - 5$.

2. Case 1: $x^2 \geq 5$

   For $x^2 \geq 5$, $x \geq \sqrt{5}$ or $x \leq -\sqrt{5}$, so the inequality simplifies to:

   $\frac{x^2 + 1}{x^2 - 5} \leq 2$

   Multiply both sides by $x^2 - 5$ (which is positive in this region):

   $x^2 + 1 \leq 2(x^2 - 5)$

   Expand and simplify:

   $x^2 + 1 \leq 2x^2 - 10$

   $1 + 10 \leq 2x^2 - x^2$

   $11 \leq x^2$

   $x \geq \sqrt{11} \quad \text{or} \quad x \leq -\sqrt{11}$

3. Case 2: $x^2 < 5$

   For $x^2 < 5$, the denominator $x^2 - 5$ is negative, so the inequality becomes:

   [ \frac{x^2 + 1}{-(x^2 - 5)} \leq 2