Answer
$$2 - 3\ln 2$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^0 {\frac{{{x^3}}}{{{x^2} - 2x + 1}}} dx \cr
& {\text{Apply long division to }}\frac{{{x^3}}}{{{x^2} - 2x + 1}} \cr
& \frac{{{x^3}}}{{{x^2} - 2x + 1}} = x + 2 + \frac{{3x - 2}}{{{x^2} - 2x + 1}} \cr
& \int_{ - 1}^0 {\frac{{{x^3}}}{{{x^2} - 2x + 1}}} dx = \int_{ - 1}^0 {\left( {x + 2 + \frac{{3x - 2}}{{{x^2} - 2x + 1}}} \right)} dx \cr
& = \int_{ - 1}^0 {\left( {x + 2} \right)} dx + \int_{ - 1}^0 {\frac{{3x - 2}}{{{x^2} - 2x + 1}}} dx \cr
& \cr
& {\text{Decompose the integrand }}\frac{{3x - 2}}{{{x^2} - 2x + 1}}{\text{ into partial fractions}} \cr
& \frac{{3x - 2}}{{{x^2} - 2x + 1}} = \frac{{3x - 2}}{{{{\left( {x - 1} \right)}^2}}} \cr
& \frac{{3x - 2}}{{{{\left( {x - 1} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} \cr
& {\text{Multiply by }}{\left( {x - 1} \right)^2} \cr
& 3x - 2 = A\left( {x - 1} \right) + B \cr
& 3x - 2 = Ax - A + B \cr
& {\text{Equate the coefficients}} \cr
& A = 3,\,\,\,\,\,\,\,\, - A + B = - 2\,\,\, \to \,\,\,\,\,\,\,\,B = 1 \cr
& \cr
& {\text{Substitute }}A,B \cr
& \frac{{3x - 2}}{{{{\left( {x - 1} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} = \frac{3}{{x - 1}} + \frac{1}{{{{\left( {x - 1} \right)}^2}}} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int_{ - 1}^0 {\left( {x + 2} \right)} dx + \int_{ - 1}^0 {\frac{{3x - 2}}{{{x^2} - 2x + 1}}} dx = \int_{ - 1}^0 {\left( {x + 2 + \frac{3}{{x - 1}} + \frac{1}{{{{\left( {x - 1} \right)}^2}}}} \right)} dx \cr
& {\text{Integrating, we get}} \cr
& = \left( {\frac{{{x^2}}}{2} + 2x + 3\ln \left| {x - 1} \right| - \frac{1}{{x - 1}}} \right)_{ - 1}^0 \cr
& = \left( {\frac{{{{\left( 0 \right)}^2}}}{2} + 2\left( 0 \right) + 3\ln \left| {0 - 1} \right| - \frac{1}{{0 - 1}}} \right) - \left( {\frac{{{{\left( { - 1} \right)}^2}}}{2} - 2\left( { - 1} \right) + 3\ln \left| { - 1 - 1} \right| + \frac{1}{{ - 1 - 1}}} \right) \cr
& = 1 - \frac{1}{2} + 2 - 3\ln 2 - \frac{1}{2} \cr
& = 2 - 3\ln 2 \cr} $$
