Answer
$$ - 2 + 3\ln 2$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{{x^3}}}{{{x^2} + 2x + 1}}} dx \cr
& {\text{Apply long division to }}\frac{{{x^3}}}{{{x^2} + 2x + 1}} \cr
& \frac{{{x^3}}}{{{x^2} + 2x + 1}} = x - 2 + \frac{{3x + 2}}{{{x^2} + 2x + 1}} \cr
& \int_0^1 {\frac{{{x^3}}}{{{x^2} + 2x + 1}}} dx = \int_0^1 {\left( {x - 2 + \frac{{3x + 2}}{{{x^2} + 2x + 1}}} \right)} dx \cr
& = \int_0^1 {\left( {x - 2} \right)} dx + \int_0^1 {\frac{{3x + 2}}{{{x^2} + 2x + 1}}} dx \cr
& \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{3x + 2}}{{{x^2} + 2x + 1}} = \frac{{3x + 2}}{{{{\left( {x + 1} \right)}^2}}} \cr
& \frac{{3x + 2}}{{{{\left( {x + 1} \right)}^2}}} = \frac{A}{{x + 1}} + \frac{B}{{{{\left( {x + 1} \right)}^2}}} \cr
& {\text{Multiply by }}{\left( {x + 1} \right)^2} \cr
& 3x + 2 = A\left( {x + 1} \right) + B \cr
& 3x + 2 = Ax + A + B \cr
& {\text{Equating coefficients}} \cr
& A = 3,\,\,\,\,\,\,\,\,A + B = 2\,\,\, \to \,\,\,\,\,\,\,\,B = - 1 \cr
& \cr
& {\text{Substituting }}A,B \cr
& \frac{{3x + 2}}{{{{\left( {x + 1} \right)}^2}}} = \frac{A}{{x + 1}} + \frac{B}{{{{\left( {x + 1} \right)}^2}}} = \frac{3}{{x + 1}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int_0^1 {\left( {x - 2} \right)} dx + \int_0^1 {\frac{{3x + 2}}{{{x^2} + 2x + 1}}} dx = \int_0^1 {\left( {x - 2 + \frac{3}{{x + 1}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)} dx \cr
& {\text{Integrating, we get}} \cr
& = \left( {\frac{{{x^2}}}{2} - 2x + 3\ln \left| {x + 1} \right| + \frac{1}{{x + 1}}} \right)_0^1 \cr
& = \left( {\frac{{{{\left( 1 \right)}^2}}}{2} - 2\left( 1 \right) + 3\ln \left| {1 + 1} \right| + \frac{1}{{1 + 1}}} \right) - \left( {\frac{{{{\left( 0 \right)}^2}}}{2} - 2\left( 0 \right) + 3\ln \left| {0 + 1} \right| + \frac{1}{{0 + 1}}} \right) \cr
& = \frac{1}{2} - 2 + 3\ln 2 + \frac{1}{2} - 1 \cr
& = - 2 + 3\ln 2 \cr} $$
