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

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

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

$\displaystyle \int \frac{x}{(x-1)(x-2)}\,dx$

Use partial fractions: $\displaystyle \frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$

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

Compare coefficients: $A+B=1,\ -2A-B=0\Rightarrow A=-1,\ B=2$

$\displaystyle \int\left(-\frac{1}{x-1}+\frac{2}{x-2}\right)dx=-\log_{e}|x-1|+2\log_{e}|x-2|+C$

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