求使得 $\frac{4x^3 + 7x^2 - 13x - 2}{(x^2 + 3)(x - 1)} = A + \frac{Bx + C}{x^2 + 3} + \frac{D}{x - 1}$ 成立的 $A$、$B$、$C$ 和 $D$ 的值。

1. Formulate the basic algebraic identity by multiplying throughout by the denominator.

$4x^3 + 7x^2 - 13x - 2 = A(x^2 + 3)(x - 1) + (Bx + C)(x - 1) + D(x^2 + 3)$

2. Compare the coefficients of $x^3$ to find $A$.

Comparing coefficients of $x^3$:
$4 = A$
$A = 4$

3. Substitute $x = 1$ to eliminate $A$, $B$, and $C$, and solve for $D$.

Let $x = 1$:
$4(1)^3 + 7(1)^2 - 13(1) - 2 = D(1^2 + 3)$
$4 + 7 - 13 - 2 = 4D$
$-4 = 4D$
$D = -1$

4. Compare the coefficients of $x^2$ to find $B$.

Comparing coefficients of $x^2$:
$7 = -A + B + D$
$7 = -4 + B - 1$
$7 = B - 5$
$B = 12$

5. Substitute $x = 0$ (or compare constant terms) to find $C$.

Let $x = 0$:
$-2 = A(3)(-1) + C(-1) + D(3)$
$-2 = -3A - C + 3D$
$-2 = -3(4) - C + 3(-1)$
$-2 = -12 - C - 3$
$-2 = -15 - C$
$C = -13$