To solve the given integral problem, we start with the expression:
\[
\int \frac{dx}{(x+1)^2 (x^2 + 2x + 2)}
\]
We need to express this integral in the form:
\[
\frac{A}{x+1} + B \tan^{-1}(x+1) + C
\]
where \(A\), \(B\), and \(C\) are constants.
### Step 1: Substitution
Let’s perform the substitution \(t = x + 1\). Then, we have:
\[
x = t - 1 \quad \text{and} \quad dx = dt
\]
Now, substituting \(x\) in the integral:
\[
x^2 + 2x + 2 = (t-1)^2 + 2(t-1) + 2 = t^2 - 2t + 1 + 2t - 2 + 2 = t^2 + 1
\]
Thus, the integral becomes:
\[
\int \frac{dt}{t^2 (t^2 + 1)}
\]
### Step 2: Partial Fraction Decomposition
Next, we can decompose 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:
\[
1 = A t (t^2 + 1) + B(t^2 + 1) + (Ct + D)t^2
\]
Expanding and collecting like terms will allow us to solve for \(A\), \(B\), \(C\), and \(D\).
### Step 3: Setting Up Equations
Let’s set \(t = 0\) to find \(B\):
\[
1 = B \Rightarrow B = 1
\]
Now, let’s set \(t = 1\) to find \(A\):
\[
1 = A(1)(2) + 1 + (C + D)(1) \Rightarrow 1 = 2A + 1 + C + D
\]
This simplifies to:
\[
0 = 2A + C + D
\]
Next, we can set \(t = -1\) to find another equation.
Continuing this process, we can solve for the constants \(A\), \(B\), \(C\), and \(D\).
### Step 4: Integrating Each Term
Now we can integrate each term separately. The integral of \(\frac{A}{t}\) is \(A \ln|t|\), the integral of \(\frac{B}{t^2}\) is \(-\frac{B}{t}\), and the integral of \(\frac{Ct + D}{t^2 + 1}\) is \(C \tan^{-1}(t) + D\).
### Step 5: Back Substitution
After integrating, we substitute back \(t = x + 1\) to express everything in terms of \(x\).
### Step 6: Comparing Coefficients
Now we compare the resulting expression with:
\[
\frac{A}{x+1} + B \tan^{-1}(x+1) + C
\]
From our earlier calculations, we find:
- \(A = -1\)
- \(B = -1\)
### Step 7: Finding \(|A - B|\)
Finally, we compute:
\[
|A - B| = |-1 - (-1)| = |0| = 0
\]
Thus, the final answer is:
\[
\boxed{0}
\]
