If `(2x^(3)+1)/((x-1)(x+1)(x+2))=A+(B)/(x-1)+(C)/(x+1)+(D)/(x+2)`, then find the value of `A+B+C+D`.

Answer

To solve the equation

\[ \frac{2x^3 + 1}{(x - 1)(x + 1)(x + 2)} = A + \frac{B}{x - 1} + \frac{C}{x + 1} + \frac{D}{x + 2}, \]

we will first multiply both sides by the denominator \((x - 1)(x + 1)(x + 2)\) to eliminate the fractions.

### Step 1: Multiply both sides by the denominator

\[ 2x^3 + 1 = A(x - 1)(x + 1)(x + 2) + B(x + 1)(x + 2) + C(x - 1)(x + 2) + D(x - 1)(x + 1). \]

### Step 2: Expand the right-hand side

Calculating each term:
- \(A(x - 1)(x + 1)(x + 2) = A(x^2 - 1)(x + 2) = A(x^3 + 2x^2 - x - 2)\)
- \(B(x + 1)(x + 2) = B(x^2 + 3x + 2)\)
- \(C(x - 1)(x + 2) = C(x^2 + x - 2)\)
- \(D(x - 1)(x + 1) = D(x^2 - 1)\)

Combining these, we have:

\[ 2x^3 + 1 = Ax^3 + (2A + B + C + D)x^2 + (-A + 3B + C)x + (-2A + 2B - 2C - D). \]

### Step 3: Compare coefficients

Now we will compare the coefficients of \(x^3\), \(x^2\), \(x\), and the constant term on both sides.

1. **Coefficient of \(x^3\)**:
\[ A = 2. \]

2. **Coefficient of \(x^2\)**:
\[ 2A + B + C + D = 0. \]
Substituting \(A = 2\):
\[ 2(2) + B + C + D = 0 \implies 4 + B + C + D = 0 \implies B + C + D = -4. \]

3. **Coefficient of \(x\)**:
\[ -A + 3B + C = 0. \]
Substituting \(A = 2\):
\[ -2 + 3B + C = 0 \implies 3B + C = 2. \]

4. **Constant term**:
\[ -2A + 2B - 2C - D = 1. \]
Substituting \(A = 2\):
\[ -2(2) + 2B - 2C - D = 1 \implies -4 + 2B - 2C - D = 1 \implies 2B - 2C - D = 5. \]

### Step 4: Solve the system of equations

We now have the following system of equations:
1. \(B + C + D = -4\) (Equation 1)
2. \(3B + C = 2\) (Equation 2)
3. \(2B - 2C - D = 5\) (Equation 3)

From Equation 2, we can express \(C\) in terms of \(B\):
\[ C = 2 - 3B. \]

Substituting \(C\) into Equation 1:
\[ B + (2 - 3B) + D = -4 \implies -2B + D = -6 \implies D = 2B - 6. \tag{4} \]

Now substitute \(C\) and \(D\) from Equations 2 and 4 into Equation 3:
\[ 2B - 2(2 - 3B) - (2B - 6) = 5 \implies 2B - 4 + 6B - 2B + 6 = 5 \implies 6B + 2 = 5 \implies 6B = 3 \implies B = \frac{1}{2}. \]

Now substitute \(B\) back to find \(C\) and \(D\):
\[ C = 2 - 3\left(\frac{1}{2}\right) = 2 - \frac{3}{2} = \frac{1}{2}, \]
\[ D = 2\left(\frac{1}{2}\right) - 6 = 1 - 6 = -5. \]

### Step 5: Calculate \(A + B + C + D\)

Now we have:
- \(A = 2\)
- \(B = \frac{1}{2}\)
- \(C = \frac{1}{2}\)
- \(D = -5\)

Thus,
\[ A + B + C + D = 2 + \frac{1}{2} + \frac{1}{2} - 5 = 2 + 1 - 5 = -2. \]

### Final Answer
The value of \(A + B + C + D\) is \(-2\).