Integrate the rational function: $\frac{x}{(x^{2}+1)(x-1)}$

Let $\frac{x}{(x^{2}+1)(x-1)} = \frac{Ax+B}{x^{2}+1} + \frac{C}{x-1}$
$x = (Ax+B)(x-1) + C(x^{2}+1)$
$x = Ax^{2} - Ax + Bx - B + Cx^{2} + C$
Equating the coefficients of $x^{2}$,$x$,and the constant term,we obtain:
$A+C = 0$
$-A+B = 1$
$-B+C = 0$
Solving these equations,we get $A = -\frac{1}{2}$,$B = \frac{1}{2}$,and $C = \frac{1}{2}$.
Substituting these values:
$\frac{x}{(x^{2}+1)(x-1)} = \frac{-\frac{1}{2}x + \frac{1}{2}}{x^{2}+1} + \frac{\frac{1}{2}}{x-1}$
Integrating both sides:
$\int \frac{x}{(x^{2}+1)(x-1)} dx = -\frac{1}{2} \int \frac{x}{x^{2}+1} dx + \frac{1}{2} \int \frac{1}{x^{2}+1} dx + \frac{1}{2} \int \frac{1}{x-1} dx$
$= -\frac{1}{4} \int \frac{2x}{x^{2}+1} dx + \frac{1}{2} \tan^{-1}(x) + \frac{1}{2} \log|x-1| + C$
$= -\frac{1}{4} \log(x^{2}+1) + \frac{1}{2} \tan^{-1}(x) + \frac{1}{2} \log|x-1| + C$
$= \frac{1}{2} \log|x-1| - \frac{1}{4} \log(x^{2}+1) + \frac{1}{2} \tan^{-1}(x) + C$