∫x/((x-1)(x-2))dx is equal to (where C is a constant of integration)

The correct answer is Option (2) → \(\log_e\left|\displaystyle\frac{(x-2)^2}{x-1}\right|+C\)

Splitting into partial fractions:

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

    \(x = A(x - 2) + B(x - 1)\)

Putting x = 1 and x = 2:

    \(x = 1:\ 1 = -A,\quad A = -1\)
    \(x = 2:\ 2 = B,\quad B = 2\)

So

    \( \displaystyle \int \frac{x}{(x-1)(x-2)}\,dx = \int \left(\frac{2}{x-2} - \frac{1}{x-1}\right) dx \)

    \( = 2\log_e|x - 2| - \log_e|x - 1| + C \)

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