Answer
$$ - \frac{1}{{18}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 1} {{{\left( {x - 2} \right)}^{ - 3}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_{ - \infty }^{ - 1} {{{\left( {x - 2} \right)}^{ - 3}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 1} {{{\left( {x - 2} \right)}^{ - 3}}} dx \cr
& {\text{integrating by using the power rule}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{{{{\left( {x - 2} \right)}^{ - 2}}}}{{ - 2}}} \right]_a^{ - 1} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{{{{\left( {x - 2} \right)}^2}}}} \right]_a^{ - 1} \cr
& {\text{property of integrals}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{{{{\left( {x - 2} \right)}^2}}}} \right]_{ - 1}^a \cr
& {\text{evaluate}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{{{{\left( {a - 2} \right)}^2}}} - \frac{1}{{{{\left( { - 1 - 2} \right)}^2}}}} \right] \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{{{{\left( {a - 2} \right)}^2}}} - \frac{1}{9}} \right] \cr
& {\text{evaluating the limit when }}a \to - \infty \cr
& = \frac{1}{2}\left( {\frac{1}{{{{\left( { - \infty - 2} \right)}^2}}} - \frac{1}{9}} \right) \cr
& = \frac{1}{2}\left( { - \frac{1}{9}} \right) \cr
& = - \frac{1}{{18}} \cr} $$
