If $\int \frac{x}{(x^2+1)(x-1)}dx=A\log |x^2+1|+B{\tan }^{-1}x+C\log |x-1|+d$ then A+B+C=

$\int \frac{x}{(x^2+1)(x-1)}dx$

Take $\frac{x}{(x^2+1)(x-1)}=\frac{Px+Q}{x^2+1}+\frac{R}{x-1}$

$x=(Px+Q)(x-1)+R(x^2+1)$

$compare x^{2}coefficients\Rightarrow 0=P+R\to \left(1\right)$

$compare xcoefficients\Rightarrow 1=Q-P\to \left(2\right)$

$compare cons\tan ts \Rightarrow 0=-Q+R\to \left(3\right)$

$\left(2\right)+\left(3\right)\Rightarrow 1=-P+R$

$\left(1\right)\Rightarrow O=P+R$

$\Rightarrow R=1/2$

$\Rightarrow Q=1/2,P=-1/2$

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

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

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

$=\frac{-1}{4}\log |x^2+1|+\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\log |x-1|+C$

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

$A+B+C=\frac{-1}{4}+\frac{1}{2}+\frac{1}{2}=\frac{3}{4}$