Answer
$$2\ln 2 + 10\left( {{e^{ - 0.3}} - {e^{ - 0.6}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^{ - 1} {\left( {\frac{{ - 2}}{t} + 3{e^{0.3t}}} \right)} dt \cr
& {\text{integrate by using }}\int_a^b {{e^{kx}}} dx = \left( {\frac{{{e^{kx}}}}{k}} \right)_a^b{\text{ and }}\int_a^b {\frac{1}{x}} dx = \left( {\ln \left| x \right|} \right)_a^b{\text{ }} \cr
& = - 2\left( {\ln \left| t \right|} \right)_{ - 2}^{ - 1} + 3\left( {\frac{{{e^{0.3t}}}}{{0.3}}} \right)_{ - 2}^{ - 1} \cr
& = - 2\left( {\ln \left| t \right|} \right)_{ - 2}^{ - 1} + 10\left( {{e^{0.3t}}} \right)_{ - 2}^{ - 1} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = - 2\left( {\ln \left| { - 1} \right| - \ln \left| { - 2} \right|} \right) + 10\left( {{e^{0.3\left( { - 1} \right)}} - {e^{0.3\left( { - 2} \right)}}} \right) \cr
& = - 2\left( { - \ln 2} \right) + 10\left( {{e^{ - 0.3}} - {e^{ - 0.6}}} \right) \cr
& = 2\ln 2 + 10\left( {{e^{ - 0.3}} - {e^{ - 0.6}}} \right) \cr} $$
