Answer
$\displaystyle \frac{5}{2}-\ln 6$
Work Step by Step
$x(x^{2}-3x+2)=x^{3}-3x^{2}+2x$
$\displaystyle \frac{x^{3}-4x+1}{x^{2}-3x+2}=\frac{x(x^{2}-3x+2)+3x^{2}-6x+1}{x^{2}-3x+2}$
$=x+\displaystyle \frac{3x^{2}-6x+1}{x^{2}-3x+2}=x+\frac{3(x^{2}-3x+2) +3x-5}{x^{2}-3x+2}$
$=x+3+\displaystyle \frac{3x-5}{x^{2}-3x+2}$
the denominator can be factored,
$=x+3+\displaystyle \frac{3x-5}{(x-2)(x-1)}$
$\displaystyle \frac{3x-5}{(x-2)(x-1)}=\frac{A}{x-2}+\frac{B}{x-1}$
$\displaystyle \frac{3x-5}{(x-2)(x-1)}=\frac{(A+B)x+(-A-2B)}{(x-2)(x-1)}$
$\left\{\begin{array}{llll}
A & +B & =3 & ...I\\
-A & -2B & =-5 & ...II
\end{array}\right.$
Adding the two equations $\Rightarrow B=2, A=1$
$\displaystyle \int_{-1}^{0}\frac{x^{3}-4x+1}{x^{2}-3x+2}dx=\int(x+3+\frac{1}{x-2}+\frac{2}{x-1})dx$
$= \left[\frac{x^{2}}{2}+3x+\ln|x-2|+2\ln|x-1|\right]_{-1}^{0}$
$=(\displaystyle \ln 2)-(\frac{1}{2}-3+\ln 3+2\ln 2)$
$=\displaystyle \frac{5}{2}-\ln 2-\ln 3$
$=\displaystyle \frac{5}{2}-(\ln 2+\ln 3)$
$=\displaystyle \frac{5}{2}-\ln 6$
