If `int(dx)/((x+1)^(2)(x^(2)+2x+2))=(A)/(x+1)+B tan^(-1)(x+1)+C,` where A and B are constants and C is the constant of integration, then `|A-B|` is equal to

Answer

To solve the integral

\[ \int \frac{dx}{(x+1)^2 (x^2 + 2x + 2)}, \]

we need to express it in the form

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

where \(A\) and \(B\) are constants, and \(C\) is the constant of integration. We will find the values of \(A\) and \(B\) and then calculate \(|A - B|\).

### Step 1: Simplify the Denominator

The expression \(x^2 + 2x + 2\) can be rewritten as:

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

Thus, the integral becomes:

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

### Step 2: Substitution

Let \(t = x + 1\). Then, \(dx = dt\) and the integral transforms to:

\[ \int \frac{dt}{t^2 (t^2 + 1)}. \]

### Step 3: Partial Fraction Decomposition

We can express the integrand using partial fractions:

\[ \frac{1}{t^2 (t^2 + 1)} = \frac{A}{t} + \frac{B}{t^2} + \frac{Ct + D}{t^2 + 1}. \]

Multiplying through by the denominator \(t^2(t^2 + 1)\) gives:

\[ 1 = A t (t^2 + 1) + B (t^2 + 1) + (Ct + D) t^2. \]

### Step 4: Expand and Collect Terms

Expanding the right-hand side:

\[ 1 = At^3 + At + Bt^2 + B + Ct^3 + Dt^2. \]

Combining like terms:

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

### Step 5: Set Up the System of Equations

Now we can set up the equations by comparing coefficients:

1. \(A + C = 0\) (coefficient of \(t^3\))
2. \(B + D = 0\) (coefficient of \(t^2\))
3. \(A = 0\) (coefficient of \(t\))
4. \(B = 1\) (constant term)

From equation 3, we have \(A = 0\). Substituting \(A = 0\) into equation 1 gives \(C = 0\). From equation 4, we have \(B = 1\). Substituting \(B = 1\) into equation 2 gives \(D = -1\).

### Step 6: Conclusion

Now we have:

- \(A = 0\)
- \(B = 1\)

We need to find \(|A - B|\):

\[ |A - B| = |0 - 1| = 1. \]

Thus, the final answer is:

\[ \boxed{1}. \]