Answer
$\frac{5}{2} - \ln \left( 6 \right)$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^0 {\frac{{{x^3} - 4x + 1}}{{{x^2} - 3x + 2}}} dx \cr
& {\text{The degree of the numerator is greater than the degree of the }} \cr
& {\text{denominator}},{\text{ we first perform the long division}} \cr
& \frac{{{x^3} - 4x + 1}}{{{x^2} - 3x + 2}} = x + 3 + \frac{{3x - 5}}{{{x^2} - 3x + 2}} \cr
& {\text{The partial fraction decomposition of }}\frac{{3x - 5}}{{{x^2} - 3x + 2}}{\text{ has the}} \cr
& {\text{form}} \cr
& \frac{{3x - 5}}{{{x^2} - 3x + 2}} = \frac{{3x - 5}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \frac{A}{{x - 2}} + \frac{B}{{x - 1}} \cr
& {\text{Multiply both sides by the least common denominator}} \cr
& 3x - 5 = A\left( {x - 1} \right) + B\left( {x - 2} \right) \cr
& x = 2 \to A = 1 \cr
& x = 1 \to B = 2 \cr
& \frac{{3x - 5}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \frac{1}{{x - 2}} + \frac{2}{{x - 1}} \cr
& {\text{Therefore}} \cr
& \frac{{{x^3} - 4x + 1}}{{{x^2} - 3x + 2}} = x + 3 + \frac{1}{{x - 2}} + \frac{2}{{x - 1}} \cr
& \int_{ - 1}^0 {\frac{{{x^3} - 4x + 1}}{{{x^2} - 3x + 2}}} dx = \int_{ - 1}^0 {\left( {x + 3 + \frac{1}{{x - 2}} + \frac{2}{{x - 1}}} \right)} dx \cr
& {\text{Intergating}} \cr
& = \left[ {\frac{1}{2}{x^2} + 3x + \ln \left| {x - 2} \right| + 2\ln \left| {x - 1} \right|} \right]_{ - 1}^0 \cr
& {\text{Evaluating}} \cr
& = \left[ {\frac{1}{2}{{\left( 0 \right)}^2} + 3\left( 0 \right) + \ln \left| {0 - 2} \right| + 2\ln \left| {0 - 1} \right|} \right] \cr
& - \left[ {\frac{1}{2}{{\left( { - 1} \right)}^2} + 3\left( { - 1} \right) + \ln \left| { - 1 - 2} \right| + 2\ln \left| { - 1 - 1} \right|} \right] \cr
& {\text{Simplifying}} \cr
& = \ln 2 - \left( { - \frac{5}{2} + \ln 3 + 2\ln 2} \right) \cr
& = \ln 2 + \frac{5}{2} - \ln 3 - 2\ln 2 \cr
& = \frac{5}{2} - \ln 3 - \ln 2 \cr
& = \frac{5}{2} - \ln \left( 6 \right) \cr} $$
