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

now,

\(\displaystyle \frac{x+3}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}\)

\(\displaystyle x+3=A(x-1)+Bx\)

comparing coeff. of x and constant term we have, A=-3 & B=4

so,

\(\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}=x-[\frac{-3}{x}+\frac{4}{x-1}]\)