Answer
$$6 - 3\ln 4$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\frac{{2x - 1}}{{x + 1}}} dx \cr
& {\text{by the long division}} \cr
& = \int_0^3 {\left( {2 - \frac{3}{{x + 1}}} \right)} dx \cr
& {\text{integrate}} \cr
& = \left. {\left( {2x - 3\ln \left| {x + 1} \right|} \right)} \right|_0^3 \cr
& {\text{using the fundamental theorem}} \cr
& = \left( {2\left( 3 \right) - 3\ln \left| {3 + 1} \right|} \right) - \left( {2\left( 0 \right) - 3\ln \left| {0 + 1} \right|} \right) \cr
& {\text{simplify}} \cr
& = \left( {6 - 3\ln \left| 4 \right|} \right) - \left( {0 - 3\ln \left| 1 \right|} \right) \cr
& = 6 - 3\ln 4 \cr} $$
