Show that

Let $\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C}{(x-1)}$

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

$x=A x^{2}-A x+B x-B+C x^{2}+C$

Equating the coefficients of x2, x, and constant term, we obtain

$A+C=0$

$-A+B=1$

$-B+C=0$

On solving these equations, we obtain

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

From equation (1), we obtain

$\therefore \frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{\left(-\frac{1}{2} x+\frac{1}{2}\right)}{x^{2}+1}+\frac{\frac{1}{2}}{(x-1)}$

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

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

Consider $\int \frac{2 x}{x^{2}+1} d x$, let $\left(x^{2}+1\right)=t \Rightarrow 2 x d x=d t$

$\Rightarrow \int \frac{2 x}{x^{2}+1} d x=\int \frac{d t}{t}=\log |t|=\log \left|x^{2}+1\right|$

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

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