To solve the equation \( 4x - 8 + \frac{4 - 2x}{x^2 + 1} = 0 \), let's follow these steps:

1. **Combine like terms and simplify the equation**:
\[
4x - 8 + \frac{4 - 2x}{x^2 + 1} = 0
\]

2. **Isolate the fraction**:
\[
4x - 8 = -\frac{4 - 2x}{x^2 + 1}
\]

3. **Multiply both sides by \( x^2 + 1 \) to eliminate the fraction**:
\[
(4x - 8)(x^2 + 1) = -(4 - 2x)
\]

4. **Expand both sides**:
\[
4x(x^2 + 1) - 8(x^2 + 1) = -4 + 2x
\]
\[
4x^3 + 4x - 8x^2 - 8 = -4 + 2x
\]

5. **Move all terms to one side to set the equation to 0**:
\[
4x^3 - 8x^2 + 4x - 8 + 4 - 2x = 0
\]
\[
4x^3 - 8x^2 + 2x - 4 = 0
\]

6. **Factor the polynomial if possible**:
Let's try to factor by grouping:
\[
4x^3 - 8x^2 + 2x - 4 = 0
\]
Group the terms:
\[
(4x^3 - 8x^2) + (2x - 4) = 0
\]
Factor out the common factors in each group:
\[
4x^2(x - 2) + 2(x - 2) = 0
\]
Factor out the common binomial factor \((x - 2)\):
\[
(x - 2)(4x^2 + 2) = 0
\]

7. **Solve for \( x \)**:
Set each factor to zero:
\[
x - 2 = 0 \quad \text{or} \quad 4x^2 + 2 = 0
\]
Solving the first equation:
\[
x = 2
\]
Solving the second equation:
\[
4x^2 + 2 = 0
\]
\[
4x^2 = -2
\]
\[
x^2 = -\frac{1}{2}
\]
Since \( x^2 = -\frac{1}{2} \) has no real solutions (as the square of a real number cannot be negative), we discard this solution.

Therefore, the only real solution is:
\[
x = 2
\]