Answer
$$ - \frac{{3 + 3\ln 2}}{2}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^{\ln 2} {\frac{{3t}}{{{e^t}}}} dt \cr
& = \int_{ - 1}^{\ln 2} {3t{e^{ - t}}} dt \cr
& {\text{substitute }}u = 3t,{\text{ }}du = 3dt \cr
& dv = {e^{ - t}}dt,{\text{ }}v = - {e^{ - t}} \cr
& {\text{applying integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& = \left( {3t} \right)\left( { - {e^{ - t}}} \right) - \int {\left( { - {e^{ - t}}} \right)\left( {3dt} \right)} \cr
& = - 3t{e^{ - t}} + 3\int {{e^{ - t}}dt} \cr
& = - 3t{e^{ - t}} - 3{e^{ - t}} + C \cr
& {\text{then}} \cr
& \int_{ - 1}^{\ln 2} {3t{e^{ - t}}} dt = \left( {\left[ { - 3t{e^{ - t}} - 3{e^{ - t}}} \right]} \right)_{ - 1}^{\ln 2} \cr
& {\text{evaluate limits}} \cr
& = \left( { - 3\left( {\ln 2} \right){e^{ - \ln 2}} - 3{e^{ - \ln 2}}} \right) - \left( { - 3\left( { - 1} \right){e^{ - \left( { - 1} \right)}} - 3{e^{ - \left( { - 1} \right)}}} \right) \cr
& {\text{simplify}} \cr
& = - 3\left( {\ln 2} \right)\left( {\frac{1}{2}} \right) - \frac{3}{2} - 3e + 3e \cr
& = - \frac{3}{2} - \frac{{3\ln 2}}{2} \cr
& = - \frac{{3 + 3\ln 2}}{2} \cr} $$
