Answer
$$\ln \sqrt 3 $$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^0 {\frac{{{e^x}}}{{3 - 2{e^x}}}} dx \cr
& \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_{ - \infty }^b {f\left( x \right)} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \int_{ - \infty }^0 {\frac{{{e^x}}}{{3 - 2{e^x}}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{{e^x}}}{{3 - 2{e^x}}}dx} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{ - 2{e^x}}}{{3 - 2{e^x}}}dx} \cr
& \cr
& {\text{Integrate}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\ln \left| {3 - 2{e^x}} \right|} \right]_a^0 \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\ln \left| {3 - 2{e^{\left( 0 \right)}}} \right| - \ln \left| {3 - 2{e^a}} \right|} \right] \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ { - \ln \left| {3 - 2{e^a}} \right|} \right] \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\ln \left| {3 - 2{e^a}} \right|} \right] \cr
& \cr
& {\text{calculate the limit when }}a \to - \infty \cr
& = \frac{1}{2}\left[ {\ln \left| {3 - 2{e^{\left( { - \infty } \right)}}} \right|} \right] \cr
& = \frac{1}{2}\left[ {\ln \left| {3 - 0} \right|} \right] \cr
& = \frac{1}{2}\ln 3 \cr
& = \ln \sqrt 3 \cr
& {\text{then}}{\text{,}} \cr
& {\text{The integral converges to }}\ln \sqrt 3 \cr} $$
