To solve the equation

\[
\frac{2x - 1}{2x + 1} - \frac{4}{1 - 4x^2} = \frac{2x + 1}{2x - 1},
\]

we first need to simplify and combine the fractions. Let's start by rewriting the second term \(\frac{4}{1 - 4x^2}\).

Notice that \(1 - 4x^2\) can be factored as a difference of squares:

\[
1 - 4x^2 = (1 - 2x)(1 + 2x).
\]

So, the term \(\frac{4}{1 - 4x^2}\) becomes:

\[
\frac{4}{(1 - 2x)(1 + 2x)}.
\]

Now, let's find a common denominator for the fractions on the left-hand side of the equation. The common denominator for \(\frac{2x - 1}{2x + 1}\) and \(\frac{4}{(1 - 2x)(1 + 2x)}\) is \((2x + 1)(1 - 2x)(1 + 2x)\).

Rewrite each fraction with this common denominator:

\[
\frac{2x - 1}{2x + 1} = \frac{(2x - 1)(1 - 2x)(1 + 2x)}{(2x + 1)(1 - 2x)(1 + 2x)},
\]

\[
\frac{4}{(1 - 2x)(1 + 2x)} = \frac{4(2x + 1)}{(2x + 1)(1 - 2x)(1 + 2x)}.
\]

Now, combine these fractions:

\[
\frac{(2x - 1)(1 - 2x)(1 + 2x) - 4(2x + 1)}{(2x + 1)(1 - 2x)(1 + 2x)}.
\]

Simplify the numerator:

\[
(2x - 1)(1 - 2x)(1 + 2x) = (2x - 1)(1 - 4x^2).
\]

So, the equation becomes:

\[
\frac{(2x - 1)(1 - 4x^2) - 4(2x + 1)}{(2x + 1)(1 - 2x)(1 + 2x)} = \frac{2x + 1}{2x - 1}.
\]

Next, simplify the numerator:

\[
(2x - 1)(1 - 4x^2) = 2x - 1 - 8x^3 + 4x^2,
\]

\[
- 4(2x + 1) = -8x - 4.
\]

Combine these:

\[
2x - 1 - 8x^3 + 4x^2 - 8x - 4 = -8x^3 + 4x^2 - 6x - 5.
\]

So, the equation becomes:

\[
\frac{-8x^3 + 4x^2 - 6x - 5}{(2x + 1)(1 - 2x)(1 + 2x)} = \frac{2x + 1}{2x - 1}.
\]

Cross-multiply to solve for \(x\):

\[
(-8x^3 + 4x^2 - 6x - 5)(2x - 1) = (2x + 1)(2x + 1)(1 - 2x)(1 + 2x).
\]

Simplify the right-hand side:

\[
(2x + 1)^2(1 - 4x^2) = (4x^2 + 4x + 1)(1 - 4x^2).
\]

Expand:

\[
(4x^2 + 4x + 1)(1 - 4x^2) = 4x^2 - 16x^4 + 4x - 16x^3 + 1 - 4x^2 = -16x^4 - 16x^3 + 4x + 1.
\]

Equate the simplified forms:

\[
-8x^3 + 4x^2 - 6x - 5 = -16x^4 - 16x^3 + 4x + 1.
\]

This is a polynomial equation. Solving it involves finding the roots of the polynomial, which can be complex. However, you can use numerical methods or factorization techniques to find the exact values of \(x\).

For simplicity, let's check for possible rational roots using the Rational Root Theorem or numerical methods.