Answer
$$ - \frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^0 {\frac{{dx}}{{{{\left( {2x - 1} \right)}^3}}}} \cr
& \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_{ - \infty }^b {f\left( x \right)} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \,\int_{ - \infty }^0 {\frac{{dx}}{{{{\left( {2x - 1} \right)}^3}}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{dx}}{{{{\left( {2x - 1} \right)}^3}}}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{2dx}}{{{{\left( {2x - 1} \right)}^3}}}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{{\left( {2x - 1} \right)}^{ - 3}}\left( 2 \right)dx} \cr
& \cr
& {\text{integrate by tables using the formula }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}}} + C \cr
& = \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{{{{\left( {2x - 1} \right)}^{ - 2}}}}{{ - 2}}} \right]_a^0 \cr
& = - \frac{1}{4}\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{{{\left( {2x - 1} \right)}^2}}}} \right]_a^0 \cr
& = - \frac{1}{4}\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{{{\left( {2\left( 0 \right) - 1} \right)}^2}}} - \frac{1}{{{{\left( {2a - 1} \right)}^2}}}} \right] \cr
& = - \frac{1}{4}\mathop {\lim }\limits_{b \to + \infty } \left[ {1 - \frac{1}{{{{\left( {2a - 1} \right)}^2}}}} \right] \cr
& \cr
& {\text{calculate the limit when }}a \to - \infty \cr
& = - \frac{1}{4}\left[ {1 - \frac{1}{{{{\left( {2\left( { - \infty } \right) - 1} \right)}^2}}}} \right] \cr
& = - \frac{1}{4}\left( {1 - \frac{1}{\infty }} \right) \cr
& = - \frac{1}{4} \cr
& {\text{then}}{\text{,}} \cr
& {\text{The integral converges to }}\, - \frac{1}{4} \cr} $$
