Answer
$$ - \frac{1}{3}\ln 7$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\frac{{dx}}{{3x - 4}}} \cr
& {\text{Use substitution:}}\cr
&{\text{Let }}u = 3x - 4,{\text{ so that }}du = 3dx \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = 3\left( 1 \right) - 4 = - 1 \cr
& \,\,\,\,\,\,{\text{If }}x = - 1,{\text{ then }}u = 3\left( { - 1} \right) - 4 = - 7 \cr
& {\text{Then}} \cr
& \int_{ - 1}^1 {\frac{{dx}}{{3x - 4}}} = \int_{ - 7}^{ - 1} {\frac{{\left( {1/3} \right)du}}{u}} \cr
& = \frac{1}{3}\int_{ - 7}^{ - 1} {\frac{1}{u}} du \cr
& {\text{integrate}} \cr
& = \frac{1}{3}\left( {\ln \left| {} \right|} \right)_{ - 7}^{ - 1} \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{1}{3}\left( {\ln \left| { - 1} \right| - \ln \left| { - 7} \right|} \right) \cr
& {\text{simplifying, we get:}} \cr
& = \frac{1}{3}\left( {0 - \ln 7} \right) \cr
& = - \frac{1}{3}\ln 7 \cr} $$
