Answer
$$ - \frac{1}{{24}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 1} {\frac{{dx}}{{{{\left( {x - 1} \right)}^4}}}} \cr
& {\text{definition of improper integral}} \cr
& \int_{ - \infty }^{ - 1} {\frac{{dx}}{{{{\left( {x - 1} \right)}^4}}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 1} {\frac{{dx}}{{{{\left( {x - 1} \right)}^4}}}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \int_a^{ - 1} {{{\left( {x - 1} \right)}^{ - 4}}} dx \cr
& {\text{evaluate the integral}} \cr
& = - \mathop {\lim }\limits_{a \to - \infty } \left. {\left( {\frac{{{{\left( {x - 1} \right)}^{ - 3}}}}{{ - 3}}} \right)} \right|_a^{ - 1} \cr
& = \frac{1}{3}\mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{{{{\left( { - 1 - 1} \right)}^3}}} - \frac{1}{{{{\left( {a - 1} \right)}^3}}}} \right) \cr
& {\text{simplify}} \cr
& = \frac{1}{3}\mathop {\lim }\limits_{a \to - \infty } \left( { - \frac{1}{8} - \frac{1}{{{{\left( {a - 1} \right)}^3}}}} \right) \cr
& {\text{evaluate the limit}} \cr
& = \frac{1}{3}\left( { - \frac{1}{8} - \frac{1}{{{{\left( { - \infty - 1} \right)}^3}}}} \right) \cr
& = \frac{1}{3}\left( { - \frac{1}{8} - \frac{1}{0}} \right) \cr
& = - \frac{1}{{24}} \cr} $$
