`int(xdx)/((x-1)(x-2))` is equals to

Options

A. `log|((x-1)^(2))/(x-2)|+C`
B. `log|((x-2)^(2))/(x-1)|+C`
C. `log|((x-1)/(x-2))^(2)|+C`
D. `log|(x-1)(x-2)|+C`

Explanation

To solve the integral \(\int \frac{x \, dx}{(x-1)(x-2)}\), we will use the method of partial fractions. Here’s a step-by-step solution:

### Step 1: Set up the partial fraction decomposition
We want to express \(\frac{x}{(x-1)(x-2)}\) in the form:
\[ \frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \]
where \(A\) and \(B\) are constants to be determined.

**Hint:** Remember that we can find \(A\) and \(B\) by multiplying both sides by the denominator \((x-1)(x-2)\) and simplifying.

### Step 2: Multiply through by the denominator
Multiplying both sides by \((x-1)(x-2)\) gives:
\[ x = A(x-2) + B(x-1) \]

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

### Step 4: Set up equations for coefficients
Now, we can equate the coefficients from both sides:
1. For \(x\): \(A + B = 1\)
2. For the constant term: \(-2A - B = 0\)

**Hint:** You have a system of linear equations to solve for \(A\) and \(B\).

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

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

### Step 7: Integrate each term
Now we can integrate:
\[ \int \frac{x \, dx}{(x-1)(x-2)} = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx \]
This gives us:
\[ = -\int \frac{1}{x-1} \, dx + 2\int \frac{1}{x-2} \, dx \]

### Step 8: Solve the integrals
The integrals can be solved as follows:
\[ -\ln |x-1| + 2\ln |x-2| + C \]
where \(C\) is the constant of integration.

### Step 9: Combine the logarithms
Using the properties of logarithms:
\[ = \ln |(x-2)^2| - \ln |x-1| + C = \ln \left| \frac{(x-2)^2}{x-1} \right| + C \]

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

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