Let, $I=\int \frac{x}{\left(x^2+1\right)\left(x-1\right)}dx$

By using integration by partial fraction method, we get

$\Rightarrow \frac{x}{\left(x^2+1\right)\left(x-1\right)} = \frac{P}{\left(x-1\right)} + \frac{Qx+R}{\left(x^2+1\right)}$

or $x = P\left(x^2+1\right) + \left(Qx+R\right)\left(x-1\right)$

or $x = \left(P+Q\right)x^2 + \left(R-Q\right)x + \left(P-R\right)$

By comparing the coefficients, we get

$P+Q=0 ......\left(1\right)$

$R-Q=1 ......\left(2\right)$

$P-R=0 ......\left(3\right)$

On solving the above three equations, we get

$P=\frac{1}{2}, Q=-\frac{1}{2}, R=\frac{1}{2}$

$\Rightarrow I = \int \frac{x}{\left(x^2+1\right)\left(x-1\right)}dx = \int \frac{{\displaystyle \frac{1}{2}}}{\left(x-1\right)}dx + \int \frac{-{\displaystyle \frac{1}{2}}x+{\displaystyle \frac{1}{2}}}{\left(x^2+1\right)}dx$

or $I = \frac{1}{2}\int \frac{1}{\left(x-1\right)}dx - \frac{1}{2}\int \frac{x}{\left(x^2+1\right)}dx + \frac{1}{2}\int \frac{1}{\left(x^2+1\right)}dx$

Let $I_1 = -\frac{1}{2}\int \frac{x}{\left(x^2+1\right)}dx$

Here we take $\left(x^2+1\right)=t$, then $xdx=\frac{dt}{2}$

$\Rightarrow I_1 = -\frac{1}{4}\int \frac{1}{t}dt$

or $I_1 = -\frac{1}{4}\log \left|t\right|$

$\Rightarrow I = \frac{1}{2}\log \left|x-1\right| - \frac{1}{4}\log \left|x^2+1\right| + \frac{1}{2}{\tan }^{-1}x + c$

$\Rightarrow A=-\frac{1}{4}, B=\frac{1}{2}, C=\frac{1}{2}$

Thus, $A+B+C = \frac{3}{4}$