Integration of rational functions
`int(dx)/((x^(2)-4x+4)(x^(2)-4x+5))`

Answer

To solve the integral

\[ \int \frac{dx}{(x^2 - 4x + 4)(x^2 - 4x + 5)}, \]

we will follow these steps:

### Step 1: Factor the Quadratics

First, we notice that \(x^2 - 4x + 4\) can be factored as \((x - 2)^2\). The second quadratic \(x^2 - 4x + 5\) cannot be factored further over the reals, but we can complete the square:

\[ x^2 - 4x + 5 = (x - 2)^2 + 1. \]

Thus, we can rewrite the integral as:

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

### Step 2: Partial Fraction Decomposition

Next, we will perform partial fraction decomposition on the integrand:

\[ \frac{1}{(x - 2)^2 \left((x - 2)^2 + 1\right)} = \frac{A}{(x - 2)} + \frac{B}{(x - 2)^2} + \frac{Cx + D}{(x - 2)^2 + 1}. \]

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

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

### Step 3: Solve for Coefficients

To find the coefficients \(A\), \(B\), \(C\), and \(D\), we can substitute convenient values for \(x\) or equate coefficients.

1. Set \(x = 2\):
\[ 1 = B(1) \implies B = 1. \]

2. Now, substituting \(B\) back into the equation and simplifying will help us find \(A\), \(C\), and \(D\).

After some algebra, we find:

\[ A = 0, \quad C = 0, \quad D = -1. \]

Thus, we have:

\[ \frac{1}{(x - 2)^2((x - 2)^2 + 1)} = \frac{0}{(x - 2)} + \frac{1}{(x - 2)^2} - \frac{1}{(x - 2)^2 + 1}. \]

### Step 4: Integrate Each Term

Now we can integrate each term separately:

1. \(\int \frac{1}{(x - 2)^2} \, dx = -\frac{1}{x - 2} + C_1\).
2. \(\int \frac{1}{(x - 2)^2 + 1} \, dx = \tan^{-1}(x - 2) + C_2\).

Combining these results, we have:

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

### Final Answer

Thus, the final result for the integral is:

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