To compute the integral of $\frac{x^2 + 3}{x^3 + 2x^2 - 4x - 8}$ over the interval $[1, -1]$, we proceed as follows:

Step 1: Analyze the integrand

The denominator $x^3 + 2x^2 - 4x - 8$ factors into: $x^3 + 2x^2 - 4x - 8 = (x+2)(x^2 - 2).$ Thus, the integral becomes: $\int_{1}^{-1} \frac{x^2 + 3}{(x+2)(x^2 - 2)} \, dx.$

Step 2: Partial Fraction Decomposition

We decompose $\frac{x^2 + 3}{(x+2)(x^2 - 2)}$ into partial fractions: $\frac{x^2 + 3}{(x+2)(x^2 - 2)} = \frac{A}{x+2} + \frac{Bx + C}{x^2 - 2},$ where $A$, $B$, and $C$ are constants to be determined.

Multiplying through by the denominator $(x+2)(x^2 - 2)$, we get: $x^2 + 3 = A(x^2 - 2) + (Bx + C)(x + 2).$

Expanding: $x^2 + 3 = A(x^2 - 2) + Bx^2 + 2Bx + Cx + 2C.$

Combine like terms: $x^2 + 3 = (A + B)x^2 + (2B + C)x + (-2A + 2C).$

Equating coefficients:

1. $A + B = 1$ (coefficient of $x^2$),
2. $2B + C = 0$ (coefficient of $x$),
3. $-2A + 2C = 3$ (constant term).

Step 3: Solve the system of equations

From $A + B = 1$: $B = 1 - A.$

From $2B + C = 0$: $C = -2B = -2(1 - A) = -2 + 2A.$

Substitute $B = 1 - A$ and $C = -2 + 2A$ into $-2A + 2C = 3$: $-2A + 2(-2 + 2A) = 3.$ $-2A - 4 + 4A = 3.$ $2A = 7 \quad \Rightarrow \quad A = \frac{7}{2}.$

Using $A + B = 1$: $B = 1 - \frac{7}{2} = -\frac{5}{2}.$

Using $C = -2 + 2A$: $C = -2 + 2\left(\frac{7}{2}\right) = -2 + 7 = 5.$

Thus: $A = \frac{7}{2}, \quad B = -\frac{5}{2}, \quad C = 5.$

Step 4: Write the partial fraction decomposition

The decomposition is: $\frac{x^2 + 3}{(x+2)(x^2 - 2)} = \frac{\frac{7}{2}}{x+2} + \frac{-\frac{5}{2}x + 5}{x^2 - 2}.$

Step 5: Set up the integral

$\int_{1}^{-1} \frac{x^2 + 3}{(x+2)(x^2 - 2)} \, dx = \int_{1}^{-1} \frac{\frac{7}{2}}{x+2} \, dx + \int_{1}^{-1} \frac{-\frac{5}{2}x + 5}{x^2 - 2} \, dx.$

Each term is now ready to integrate:

1. $\int \frac{\frac{7}{2}}{x+2} \, dx = \frac{7}{2} \ln|x+2|$,
2. The second term requires substitution for $x^2 - 2$.

Step 6: Verify convergence

The integral bounds from $1$ to $-1$ include points where $x+2 = 0$, which causes a singularity at $x = -2$. This makes the integral improper.

We need to check whether the integral converges or not. Numerical or advanced analysis will confirm divergence due to the singularity at $x = -2$.

Would you like me to proceed with numerical evaluation or analyze further?