We are tasked with integrating the following expression:

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

### Step 1: Split the integral

We can break the expression into two separate integrals:

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

### Step 2: Solve the first integral

The integral of \( x \) is straightforward:

\[
\int x \, dx = \frac{x^2}{2}
\]

### Step 3: Solve the second integral using partial fractions

To integrate \( \frac{1}{(x - 1)(x - 2)} \), we can decompose it into partial fractions:

\[
\frac{1}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}
\]

Multiplying both sides by \( (x - 1)(x - 2) \) gives:

\[
1 = A(x - 2) + B(x - 1)
\]

Expanding:

\[
1 = A(x) - 2A + B(x) - B
\]

\[
1 = (A + B)x - (2A + B)
\]

Now, equate the coefficients of \( x \) and the constant term on both sides:

1. \( A + B = 0 \)
2. \( -2A - B = 1 \)

From the first equation, \( A = -B \). Substituting this into the second equation:

\[
-2(-B) - B = 1
\]

\[
2B - B = 1 \implies B = 1
\]

Since \( B = 1 \), \( A = -1 \).

Thus, we can rewrite the fraction as:

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

### Step 4: Integrate the partial fractions

Now, we can integrate:

\[
\int \frac{-1}{x - 1} \, dx + \int \frac{1}{x - 2} \, dx
\]

These are simple logarithmic integrals:

\[
- \ln |x - 1| + \ln |x - 2|
\]

### Step 5: Combine the results

Combining everything, the integral is:

\[
\frac{x^2}{2} - \ln |x - 1| + \ln |x - 2| + C
\]

We can also combine the logarithmic terms:

\[
\frac{x^2}{2} + \ln \left| \frac{x - 2}{x - 1} \right| + C
\]

This is the final solution.

### Do you want further details on any step, or have any questions?

Here are 5 related questions:
1. How do you find partial fraction decomposition for more complex denominators?
2. What is the rule for integrating logarithmic functions?
3. Can you use substitution to solve integrals like this?
4. How do you handle improper integrals with similar forms?
5. What are the conditions under which partial fractions can be applied?

**Tip:** When integrating rational functions, partial fractions are extremely useful for simplifying the integrand into a sum of easier integrals.

This detailed guide explains how to integrate the expression x + 1/(x-1)(x-2) using partial fraction decomposition. Follow step-by-step instructions that cover key calculus concepts like the integration of rational functions and logarithmic integrals. Suitable for undergraduate students.

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