Answer
$$ - \frac{1}{4}{e^{ - 2x}}\left( {2x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {x{e^{ - 2x}}} dx,\,\,\,\,u = - 2x \cr
& {\text{Using the given substitution}} \cr
& u = - 2x,\,\,\,\,\,\,\,du = - 2dx,\,\,\,\,\,dx = - \frac{{du}}{2}\, \cr
& {\text{write in terms of }}u \cr
& \int {x{e^{ - 2x}}} dx = - \frac{1}{2}\int {\left( { - 2x} \right){e^{ - 2x}}} dx = \frac{{ - 1}}{2}\int {\left( { - 2x} \right){e^{ - 2x}}} \left( { - \frac{{du}}{2}\,} \right) \cr
& = \frac{1}{4}\int {u{e^u}} du \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 51}} \cr
& \left( {51} \right):\,\,\,\,\,\,\int {u{e^u}du} = {e^u}\left( {u - 1} \right) + C \cr
& \frac{1}{4}\int {u{e^u}} du = \frac{1}{4}{e^u}\left( {u - 1} \right) + C \cr
& {\text{write in terms of }}x{\text{, and replace }} - 2x{\text{ for }}u \cr
& = \frac{1}{4}{e^{ - 2x}}\left( { - 2x - 1} \right) + C \cr
& = - \frac{1}{4}{e^{ - 2x}}\left( {2x + 1} \right) + C \cr} $$
