`int (x dx)/((x-1)(x-2)) ` equals:

Answer

To solve the integral \( \int \frac{x \, dx}{(x-1)(x-2)} \), we will use the method of partial fractions.

### Step 1: Set up the partial fraction decomposition
We can express the integrand as:
\[ \frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \]
where \( A \) and \( B \) are constants that we need to determine.

### Step 2: Clear the denominators
Multiply both sides by \( (x-1)(x-2) \):
\[ x = A(x-2) + B(x-1) \]

### Step 3: Expand the right-hand side
Expanding the right side gives:
\[ x = Ax - 2A + Bx - B \]
Combining like terms:
\[ x = (A + B)x - (2A + B) \]

### Step 4: Set up equations for coefficients
For the equation to hold for all \( x \), the coefficients of \( x \) on both sides must be equal, and the constant terms must also be equal. This gives us the system of equations:
1. \( A + B = 1 \) (coefficient of \( x \))
2. \( -2A - B = 0 \) (constant term)

### Step 5: Solve the system of equations
From the second equation, we can express \( B \) in terms of \( A \):
\[ B = -2A \]
Substituting into the first equation:
\[ A - 2A = 1 \implies -A = 1 \implies A = -1 \]
Now substituting \( A \) back to find \( B \):
\[ B = -2(-1) = 2 \]

### Step 6: Write the partial fraction decomposition
Now we have:
\[ \frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2} \]

### Step 7: Rewrite the integral
We can now rewrite the integral:
\[ \int \frac{x \, dx}{(x-1)(x-2)} = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx \]

### Step 8: Integrate term by term
Integrating each term separately:
\[ \int \frac{-1}{x-1} \, dx + 2 \int \frac{1}{x-2} \, dx \]
This gives:
\[ - \ln |x-1| + 2 \ln |x-2| + C \]

### Step 9: Simplify the result
Using the properties of logarithms, we can combine the logarithmic terms:
\[ - \ln |x-1| + \ln |(x-2)^2| = \ln \left( \frac{(x-2)^2}{|x-1|} \right) \]

### Final Result
Thus, the final answer is:
\[ \int \frac{x \, dx}{(x-1)(x-2)} = \ln \left( \frac{(x-2)^2}{|x-1|} \right) + C \]