Answer
$$1 - \ln 3$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{{x^2} - x}}{{{x^2} + x + 1}}} dx \cr
& {\text{By long division}} \cr
& \frac{{{x^2} - x}}{{{x^2} + x + 1}} = 1 - \frac{{2x + 1}}{{{x^2} + x + 1}} \cr
& {\text{Therefore,}} \cr
& \int_0^1 {\frac{{{x^2} - x}}{{{x^2} + x + 1}}} dx = \int_0^1 {\left( {1 - \frac{{2x + 1}}{{{x^2} + x + 1}}} \right)} dx \cr
& {\text{Integrating }} \cr
& = \left[ {x - \ln \left| {{x^2} + x + 1} \right|} \right]_0^1 \cr
& = \left[ {1 - \ln \left| {{1^2} + 1 + 1} \right|} \right] - \left[ {0 - \ln \left| {{0^2} + 0 + 1} \right|} \right] \cr
& = \left[ {1 - \ln \left| 3 \right|} \right] - \left[ {0 - \ln \left| 1 \right|} \right] \cr
& = 1 - \ln 3 \cr} $$
