Answer
$${\text{The integral is divergent}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 1} {\frac{{2x - 1}}{{{x^2} - x}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{}} \cr
& \int_{ - \infty }^{ - 1} {\frac{{2x - 1}}{{{x^2} - x}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 1} {\frac{{2x - 1}}{{{x^2} - x}}} dx \cr
& \cr
& {\text{integrating }}\int {\frac{{2x - 1}}{{{x^2} - x}}} dx \cr
& {\text{set }}u = {x^2} - x{\text{ then }}\frac{{du}}{{dx}} = \left( {2x - 1} \right),\,\,\,\,\frac{{du}}{{2x - 1}} = dx \cr
& \int {\frac{{2x - 1}}{{{x^2} - x}}} dx = \int {\frac{{2x - 1}}{u}} \left( {\frac{{du}}{{2x - 1}}} \right) = \int {\frac{{du}}{u}} \cr
& = \ln \left| u \right| + C \cr
& {\text{replace }}u = {x^2} - x \cr
& = \ln \left| {{x^2} - x} \right| + C \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 1} {\frac{{2x - 1}}{{{x^2} - x}}} dx = \mathop {\lim }\limits_{a \to - \infty } \left( {\ln \left| {{x^2} - x} \right|} \right)_a^{ - 1} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\ln \left| {{{\left( { - 1} \right)}^2} - \left( { - 1} \right)} \right| - \ln \left| {{{\left( a \right)}^2} - \left( a \right)} \right|} \right) \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\ln 2 - \ln \left| {{a^2} - a} \right|} \right) \cr
& {\text{evaluating the limit when }}a \to - \infty \cr
& = \ln 2 - \ln \left| {{{\left( { - \infty } \right)}^2} - \left( { - \infty } \right)} \right| \cr
& = \infty \cr
& then \cr
& {\text{The integral is divergent}} \cr} $$
