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

Answer

To solve the equation

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

we start by simplifying the left-hand side.

### Step 1: Simplify the Denominator
The denominator \(x^2 + 2x + 1\) can be factored as:

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

Thus, we can rewrite the left-hand side:

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

### Step 2: Multiply Both Sides by \((x + 1)^2\)
To eliminate the denominator, we multiply both sides by \((x + 1)^2\):

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

### Step 3: Expand the Right-Hand Side
Now, we expand the right-hand side:

\[ A(x + 1)^2 = A(x^2 + 2x + 1) = Ax^2 + 2Ax + A, \]
\[ B(x + 1) = Bx + B, \]
\[ C = C. \]

Combining these, we get:

\[ Ax^2 + (2A + B)x + (A + B + C). \]

### Step 4: Set Up the Equation
Now we equate the coefficients from both sides:

\[ x^2 + x + 2 = Ax^2 + (2A + B)x + (A + B + C). \]

### Step 5: Compare Coefficients
From the left-hand side, we have:

- Coefficient of \(x^2\): \(1\)
- Coefficient of \(x\): \(1\)
- Constant term: \(2\)

From the right-hand side, we have:

- Coefficient of \(x^2\): \(A\)
- Coefficient of \(x\): \(2A + B\)
- Constant term: \(A + B + C\)

Setting these equal gives us the following equations:

1. \(A = 1\)
2. \(2A + B = 1\)
3. \(A + B + C = 2\)

### Step 6: Solve the Equations
From the first equation, we have:

\[ A = 1. \]

Substituting \(A = 1\) into the second equation:

\[ 2(1) + B = 1 \implies 2 + B = 1 \implies B = 1 - 2 = -1. \]

Now substituting \(A = 1\) and \(B = -1\) into the third equation:

\[ 1 - 1 + C = 2 \implies C = 2. \]

### Step 7: Find \(A + B + C\)
Now we can find \(A + B + C\):

\[ A + B + C = 1 - 1 + 2 = 2. \]

Thus, the final answer is:

\[ \boxed{2}. \]