To solve the equation

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

let's first clear the fractions by finding a common denominator. The common denominator is \(2x + 1\). Multiply every term by \(2x + 1\) to get:

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

Simplify the right-hand side:

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

Distribute on the right:

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

Simplify further:

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

Now distribute the left side:

\[ 2x^2 - 2 = 3x. \]

Rearrange the equation into a standard quadratic form:

\[ 2x^2 - 3x - 2 = 0. \]

To solve this quadratic equation, we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]

where \( a = 2 \), \( b = -3 \), and \( c = -2 \). Plug these values into the formula:

\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}. \]

Calculate the discriminant:

\[ (-3)^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25. \]

The discriminant is 25, which is a perfect square. Hence, the roots are rational:

\[ x = \frac{3 \pm \sqrt{25}}{4}. \]

\[ x = \frac{3 \pm 5}{4}. \]

This gives two solutions:

1. \( x = \frac{3 + 5}{4} = \frac{8}{4} = 2, \)
2. \( x = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac{1}{2}. \)

Thus, the solutions to the equation are \( x = 2 \) and \( x = -\frac{1}{2} \). However, substituting these values back into the original equation to ensure they are within the domain as the original equation had a denominator dependent on \( 2x + 1 \). Particularly, check that \( 2x + 1 \neq 0 \).

For \( x = 2 \):
\[ 2 \times 2 + 1 = 5 \neq 0, \text{ so } x = 2 \text{ is valid.} \]

For \( x = -\frac{1}{2} \):
\[ 2 \times \left(-\frac{1}{2}\right) + 1 = -1 + 1 = 0, \text{ so } x = -\frac{1}{2} \text{ is not valid due to division by zero.} \]

Therefore, the only valid solution is \( x = 2 \).