Step 1: Understanding the Concept:
The integrand is a rational function where the degree of the numerator is less than the degree of the denominator, and the denominator factors into distinct linear terms. This is a textbook case for integration using partial fractions.
Step 2: Key Formula or Approach:
Express the fraction as a sum of simpler fractions: $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.
Solve for constants $A$ and $B$, then integrate each term using $\int \frac{1}{x-a} dx = \log|x-a| + C$.
Step 3: Detailed Explanation:
Set up the partial fraction decomposition:
\[ \frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \]
Multiply both sides by the common denominator $(x-1)(x-2)$:
\[ x = A(x-2) + B(x-1) \]
To find $A$, let $x = 1$:
\[ 1 = A(1-2) + B(0) \implies 1 = -A \implies A = -1 \]
To find $B$, let $x = 2$:
\[ 2 = A(0) + B(2-1) \implies 2 = B \implies B = 2 \]
Now substitute $A$ and $B$ back into the integral:
\[ \int \frac{x}{(x-1)(x-2)} dx = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx \]
Integrate term by term:
\[ = -\log|x-1| + 2\log|x-2| + c \]
Use logarithm properties to combine the terms. Recall $n\log a = \log(a^n)$ and $\log a - \log b = \log(\frac{a}{b})$:
\[ = \log|x-2|^2 - \log|x-1| + c \]
Since $|x-2|^2 = (x-2)^2$, we can drop the absolute value for the squared term. For matching the options, we assume the arguments are positive in their domain:
\[ = \log(x-2)^2 - \log(x-1) + c \]
\[ = \log \left( \frac{(x-2)^2}{x-1} \right) + c \]
Step 4: Final Answer:
The evaluated integral matches option (D).