Show that

$\frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{3 x+5}{(x-1)^{2}(x+1)}$

Let $\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+1)}$

$3 x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2}$

$3 x+5=A\left(x^{2}-1\right)+B(x+1)+C\left(x^{2}+1-2 x\right)$   ...(1)

Substituting x = 1 in equation (1), we obtain

B = 4

Equating the coefficients of $x^{2}$ and $x$, we obtain

A + C = 0

B − 2C = 3

On solving, we obtain

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

$\therefore \frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}$

$\Rightarrow \int \frac{3 x+5}{(x-1)^{2}(x+1)} d x=-\frac{1}{2} \int \frac{1}{x-1} d x+4 \int \frac{1}{(x-1)^{2}} d x+\frac{1}{2} \int \frac{1}{(x+1)} d x$

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

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