$\int \frac{x}{(x-1)(x-2)} dx$ is equal to ( where $C$ is a constant of integration)

The integral to evaluate is $\int \frac{x}{(x-1)(x-2)} dx$

The fraction can be decomposed into partial fractions:

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

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Multiplying both sides by $(x-1)(x-2)$:

$x = A(x-2) + B(x-1)$

For $x = 1$:

$1 = A(1-2) + B(0)$

$1 = -A$

$A = -1$

For $x = 2$:

$2 = A(0) + B(2-1)$

$2 = B$

$B = 2$

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The fraction becomes:

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

The integral can be written as:

$\int \frac{x}{(x-1)(x-2)} dx = \int \frac{-1}{x-1} dx + \int \frac{2}{x-2} dx$

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Integrating each term:

$\int \frac{x}{(x-1)(x-2)} dx = -\ln|x-1| + 2\ln|x-2| + C$

$= -\ln|x-1| + \ln|(x-2)^2| + C$

$= \ln|(x-2)^2| - \ln|x-1| + C$

$= \ln\left|\frac{(x-2)^2}{x-1}\right| + C$

Therefore, $\int \frac{x}{(x-1)(x-2)} dx = \ln\left|\frac{(x-2)^2}{x-1}\right| + C$