Calculus, 10th Edition (Anton)

$$3\ln 3 - 2$$
\eqalign{ & \int_{ - 1}^1 {\ln \left( {x + 2} \right)} dx \cr & {\text{substitute }}u = \ln \left( {x + 2} \right),{\text{ }}du = \frac{1}{{x + 2}}dx \cr & dv = dx,{\text{ }}v = x \cr & {\text{applying integration by parts}}{\text{, we have}} \cr & \int_{ - 1}^1 {\ln \left( {x + 2} \right)} dx = \left. {\left( {x\ln \left( {x + 2} \right)} \right)} \right|_{ - 1}^1 - \int_{ - 1}^1 {\left( x \right)\left( {\frac{1}{{x + 2}}dx} \right)} \cr & \int_{ - 1}^1 {\ln \left( {x + 2} \right)} dx = \left. {\left( {x\ln \left( {x + 2} \right)} \right)} \right|_{ - 1}^1 - \int_{ - 1}^1 {\frac{x}{{x + 2}}dx} \cr & {\text{long division}} \cr & \int_{ - 1}^1 {\ln \left( {x + 2} \right)} dx = \left. {\left( {x\ln \left( {x + 2} \right)} \right)} \right|_{ - 1}^1 - \int_{ - 1}^1 {\left( {1 - \frac{2}{{x + 2}}} \right)dx} \cr & {\text{integrating}} \cr & \int_{ - 1}^1 {\ln \left( {x + 2} \right)} dx = \left. {\left( {x\ln \left( {x + 2} \right)} \right)} \right|_{ - 1}^1 - \left. {\left( {x - 2\ln \left| {x + 2} \right|} \right)} \right|_{ - 1}^1 \cr & \int_{ - 1}^1 {\ln \left( {x + 2} \right)} dx = \left. {\left( {x\ln \left( {x + 2} \right) - x + 2\ln \left| {x + 2} \right|} \right)} \right|_{ - 1}^1 \cr & {\text{evaluate limits}} \cr & = \left( {1\ln \left( {1 + 2} \right) - 1 + 2\ln \left| {1 + 2} \right|} \right) - \left( { - \ln \left( { - 1 + 2} \right) - \left( { - 1} \right) + 2\ln \left| { - 1 + 2} \right|} \right) \cr & = \ln 3 - 1 + 2\ln 3 - 1 \cr & {\text{simplify}} \cr & = 3\ln 3 - 2 \cr}