在不使用图形计算器的情况下，求解不等式 $\frac{x+2}{2x-1} < 2x+1$。

1. Solve the inequality from the first part.

$\frac{x+2}{2x-1} < 2x+1$
$\frac{x+2}{2x-1} - \frac{(2x+1)(2x-1)}{2x-1} < 0$
$\frac{x+2 - (4x^2 - 1)}{2x-1} < 0$
$\frac{-4x^2 + x + 3}{2x-1} < 0$
$\frac{4x^2 - x - 3}{2x-1} > 0$
$\frac{(4x+3)(x-1)}{2x-1} > 0$
The critical values are $x = -\frac{3}{4}$, $x = \frac{1}{2}$, and $x = 1$.
By performing a sign test, the solution is:
$-\frac{3}{4} < x < \frac{1}{2} \quad \text{or} \quad x > 1$

2. Relate the target inequality to the first part using a substitution.

Let $y = \frac{1}{x^2}$. We rewrite the left-hand side of the target inequality:
$\frac{2x^2+1}{2-x^2} = \frac{2\left(\frac{1}{y}\right)+1}{2-\frac{1}{y}} = \frac{\frac{2+y}{y}}{\frac{2y-1}{y}} = \frac{y+2}{2y-1}$
The right-hand side becomes:
$\frac{2+x^2}{x^2} = \frac{2}{x^2} + 1 = 2y + 1$
Thus, the inequality $\frac{2x^2+1}{2-x^2} < \frac{2+x^2}{x^2}$ is exactly equivalent to:
$\frac{y+2}{2y-1} < 2y+1$

3. Use the solution from the first part to set up inequalities for $x$.

Since this matches the inequality solved in the first part, the solution for $y$ is:
$-\frac{3}{4} < y < \frac{1}{2} \quad \text{or} \quad y > 1$
Substituting $y = \frac{1}{x^2}$ back into the inequalities:
$-\frac{3}{4} < \frac{1}{x^2} < \frac{1}{2} \quad \text{or} \quad \frac{1}{x^2} > 1$

4. Solve the resulting inequalities for $x$, noting that $x^2 > 0$.

Since $x^2 > 0$ for all real $x \neq 0$, the term $\frac{1}{x^2}$ is always strictly positive.
For the first inequality:
$0 < \frac{1}{x^2} < \frac{1}{2} \implies x^2 > 2$
$x < -\sqrt{2} \quad \text{or} \quad x > \sqrt{2}$
For the second inequality:
$\frac{1}{x^2} > 1 \implies x^2 < 1 \implies -1 < x < 1$
Since $x \neq 0$ (otherwise $\frac{1}{x^2}$ is undefined), this yields:
$-1 < x < 0 \quad \text{or} \quad 0 < x < 1$
Combining all parts, the exact solution is:
$x < -\sqrt{2}, \quad -1 < x < 0, \quad 0 < x < 1 \quad \text{or} \quad x > \sqrt{2}$