Answer
$\ln 9 - 4$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^3 {\frac{{4{x^2} - 7}}{{2x + 3}}} dx \cr
& {\text{Apply long division to the integrand }}\frac{{4{x^2} - 7}}{{2x + 3}}{\text{ }} \cr
& \frac{{4{x^2} - 7}}{{2x + 3}} = 2x - 3 + \frac{2}{{2x + 3}} \cr
& {\text{Then}}{\text{,}} \cr
& \cr
& \int_{ - 1}^3 {\frac{{4{x^2} - 7}}{{2x + 3}}} dx = \int_{ - 1}^3 {\left( {2x - 3 + \frac{2}{{2x + 3}}} \right)} dx \cr
& \cr
& {\text{Integrating}} \cr
& = \left( {{x^2} - 3x + \ln \left| {2x + 3} \right|} \right)_{ - 1}^3 \cr
& = \left( {{{\left( 3 \right)}^2} - 3\left( 3 \right) + \ln \left| {2\left( 3 \right) + 3} \right|} \right) - \left( {{{\left( { - 1} \right)}^2} - 3\left( { - 1} \right) + \ln \left| {2\left( { - 1} \right) + 3} \right|} \right) \cr
& {\text{Simplifying, we get}} \cr
& = \left( {9 - 9 + \ln 9} \right) - \left( {1 + 3 + \ln 1} \right) \cr
& = \ln 9 - 4 \cr} $$
