By using the substitution $y = |x|$, solve the inequality $\frac{|x| + 3}{x^2 + 1} > 1$.

1. Apply the substitution $y = |x|$ and clear the denominator to form a polynomial inequality.

Let $y = |x|$. Then $x^2 = |x|^2 = y^2$.
Substitute $y$ into the inequality:
$\frac{y + 3}{y^2 + 1} > 1$
Since $y^2 + 1 > 0$ for all real $y$, we can multiply both sides by $y^2 + 1$ without changing the inequality sign:
$y + 3 > y^2 + 1$
$y^2 - y - 2 < 0$

2. Solve the quadratic inequality for $y$.

Factorise the quadratic expression:
$(y - 2)(y + 1) < 0$
$-1 < y < 2$

3. Substitute back $y = |x|$ and solve the modulus inequality for $x$.

Substitute $y = |x|$ back into the inequality:
$-1 < |x| < 2$
Since $|x| \ge 0$ for all real $x$, the inequality $-1 < |x|$ is always true. We only need to solve:
$|x| < 2$
$-2 < x < 2$