$\cfrac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}$

Let $I = \int {\cfrac{{xdx}}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}}$

We write, $\cfrac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}} = \cfrac{A}{{x - 1}} + \cfrac{{Bx + C}}{{{x^2} + 1}}$

$\Rightarrow$ $x = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)\left( {x - 1} \right)$

….(i)
Putting $x = 1$ in $\left( i \right)$ ,

we get
$1 = A\left( {1 + 1} \right)$ $\Rightarrow$ $A = \cfrac{1}{2}$
Comparing coefficients of ${x^2}$ in (i),

we get
$0 = A + B$ $\Rightarrow$ $B = - \cfrac{1}{2}$

Comparing the constant terms,

we get
$0 = A - C$ $\Rightarrow$ $C = \cfrac{1}{2}$

$\therefore$ $I = \int {\cfrac{x}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}dx} = \int {\left[ {\cfrac{{1/2}}{{x - 1}} + \cfrac{{ - \cfrac{1}{2}x + \cfrac{1}{2}}}{{{x^2} + 1}}} \right]dx}$

$= \cfrac{1}{2}\int {\cfrac{{dx}}{{x - 1}}} - \cfrac{1}{2}\int {\cfrac{x}{{{x^2} + 1}}dx + \cfrac{1}{2}\int {\cfrac{{dx}}{{{x^2} + 1}}} }$

$= \cfrac{1}{2}\log \left( {x - 1} \right) - \cfrac{1}{4}\int {\cfrac{{2x}}{{{x^2} + 1}}} + \cfrac{1}{2}{\tan ^{ - 1}}x + C$

$= \cfrac{1}{2}\log \left( {x - 1} \right) - \cfrac{1}{4}\log \left( {{x^2} + 1} \right) + \cfrac{1}{2}{\tan ^{ - 1}}x + C$