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

1. Substitute $y = |x|$ into the inequality and simplify to a quadratic inequality.

Recognise that $x^2 = |x|^2 = y^2$. Substituting $y = |x|$ into the inequality gives:
$\frac{y+3}{y^2+1} > 1$
Since $y^2+1 > 0$ for all real $y$, we can safely 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:
$(y-2)(y+1) < 0$
Solving this yields:
$-1 < y < 2$

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

Substitute $y = |x|$ back into the solution:
$-1 < |x| < 2$
Since $|x| \ge 0$ for all real $x$, the left-hand inequality $-1 < |x|$ is always satisfied. We only need to solve the right-hand inequality:
$|x| < 2$
Using the property $|x| < a \iff -a < x < a$:
$-2 < x < 2$