Integrate the rational function: $\frac{3x+5}{x^{3}-x^{2}-x+1}$

(A) First,factor the denominator: $x^{3}-x^{2}-x+1 = x^{2}(x-1)-1(x-1) = (x^{2}-1)(x-1) = (x-1)(x+1)(x-1) = (x-1)^{2}(x+1)$.
Using partial fractions,let $\frac{3x+5}{(x-1)^{2}(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{C}{x+1}$.
Multiplying by the denominator,we get $3x+5 = A(x-1)(x+1) + B(x+1) + C(x-1)^{2}$.
Setting $x=1$,we get $3(1)+5 = B(1+1) \Rightarrow 8 = 2B \Rightarrow B=4$.
Setting $x=-1$,we get $3(-1)+5 = C(-1-1)^{2} \Rightarrow 2 = 4C \Rightarrow C=\frac{1}{2}$.
Comparing coefficients of $x^{2}$,we get $A+C=0 \Rightarrow A = -C = -\frac{1}{2}$.
Thus,$\frac{3x+5}{(x-1)^{2}(x+1)} = -\frac{1}{2(x-1)} + \frac{4}{(x-1)^{2}} + \frac{1}{2(x+1)}$.
Integrating both sides: $\int \frac{3x+5}{(x-1)^{2}(x+1)} dx = -\frac{1}{2} \int \frac{1}{x-1} dx + 4 \int (x-1)^{-2} dx + \frac{1}{2} \int \frac{1}{x+1} dx$.
$= -\frac{1}{2} \ln|x-1| - \frac{4}{x-1} + \frac{1}{2} \ln|x+1| + K$.
$= \frac{1}{2} \ln \left| \frac{x+1}{x-1} \right| - \frac{4}{x-1} + K$,where $K$ is the constant of integration.