通过换元 $y = |x|$，解不等式 $\frac{|x|+3}{x^2+1} > 1$。

1. Use the substitution $y = |x|$ to convert the rational inequality into a quadratic inequality in $y$.

Since $y = |x|$, we have $y^2 = |x|^2 = x^2$.
Substituting into the given inequality:
$\frac{y+3}{y^2+1} > 1$
Since $y^2+1 > 0$ for all real values of $y$, we can safely multiply both sides by $y^2+1$ without reversing the inequality sign:
$y + 3 > y^2 + 1$
$y^2 - y - 2 < 0$

2. Solve the quadratic inequality for $y$.

Factorising the quadratic expression:
$(y - 2)(y + 1) < 0$
Solving for $y$ gives:
$-1 < y < 2$

3. Substitute $y = |x|$ back and solve the resulting absolute value inequality for $x$.

$-1 < |x| < 2$
Since $|x| \ge 0$ for all real $x$, the left side of the inequality ($-1 < |x|$) is always satisfied.
Therefore, we only need to solve the right side:
$|x| < 2$
$-2 < x < 2$