$\int {\cfrac{{x\,dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}}$ equals

Option b is correct

: $I = \int {\cfrac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)}}} dx$

We write $\cfrac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \cfrac{A}{{x - 1}} + \cfrac{B}{{x - 2}}$

$\Rightarrow$ $x = A\left( {x - 2} \right) + B\left( {x - 1} \right)$
Comparing like terms,

we get
$A + B = 1$ and $- 2A - B = 0$
Solving,

we get $A = - 1,B = 2$
$\therefore$ $I = \int {\left( {\cfrac{{ - 1}}{{x - 1}} + \cfrac{2}{{x - 2}}} \right)dx}$

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