Answer
$\frac{5}{2}-\ln6$
Work Step by Step
$\int_{-1}^0{\frac{x^3-4x+1}{x^2-3x+2}}dx$ = $\int_{-1}^0\left({x+3+\frac{3x-5}{x^2-3x+2}}\right)dx$ = $\int_{-1}^0\left[{x+3+\frac{3x-5}{(x-1)(x-2)}}\right]dx$
$\frac{3x-5}{(x-1)(x-2)}$ = $\frac{A}{x-1}+\frac{B}{x-2}$
$3x-5$ = $A(x-2)+B(x-1)$
$A+B$ = $3$
$-2A-B$ = $-5$
$A$ = $2$, $B$ = $1$
So
$\int_{-1}^0\left[{x+3+\frac{3x-5}{(x-1)(x-2)}}\right]dx$ = $\int_{-1}^0\left[{x+3+\frac{2}{x-1}+\frac{1}{x-2}}\right]dx$ = $\left[\frac{x^2}{2}+3x+2\ln|x-1|+\ln|x-2|\right]_{-1}^0$ = $(2\ln1+\ln2)-\left(\frac{1}{2}-3+2\ln2+\ln3\right)$ =$\frac{5}{2}-\ln6$
