$\frac{3 x+5}{x^{3}-x^{2}-x+1}$
$\frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{3 x+5}{(x-1)^{2}(x+1)}$
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 $C=\frac{1}{2}$
$\therefore \frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}$
$\Rightarrow \int \frac{3 x+5}{(x-1)^{2}(x+1)} d x=-\frac{1}{2} \int \frac{1}{x-1} d x+4 \int \frac{1}{(x-1)^{2}} d x+\frac{1}{2} \int \frac{1}{(x+1)} d x$
$=-\frac{1}{2} \log |x-1|+4\left(\frac{-1}{x-1}\right)+\frac{1}{2} \log |x+1|+\mathrm{C}$
$=\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{(x-1)}+\mathrm{C}$