Answer
$$ - \frac{5}{e} + 3$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{2x + 1}}{{{e^x}}}} dx \cr
& {\text{use property of exponents }}\frac{1}{{{e^a}}} = {e^{ - a}} \cr
& \int {\left( {2x + 1} \right){e^{ - x}}} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = 2x + 1{\text{ then }}du = 2dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{ - x}}dx{\text{ then }}v = - {e^{ - x}} \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\left( {2x + 1} \right){e^{ - x}}} dx = \left( {2x + 1} \right)\left( { - {e^{ - x}}} \right) - \int {\left( { - {e^{ - x}}} \right)\left( 2 \right)} dx \cr
& \int {\left( {2x + 1} \right){e^{ - x}}} dx = - \left( {2x + 1} \right){e^{ - x}} + 2\int {{e^{ - x}}} dx \cr
& {\text{integrate using }}\int {{e^{kx}}dx} = \frac{{{e^{kx}}}}{k} + C \cr
& \int {\left( {2x + 1} \right){e^{ - x}}} dx = - \left( {2x + 1} \right){e^{ - x}} + 2\left( { - {e^{ - x}}} \right) + C \cr
& \int_0^1 {\frac{{2x + 1}}{{{e^x}}}} dx = - \left( {2x + 1} \right){e^{ - x}} - 2{e^{ - x}} + C \cr
& \cr
& {\text{Now find the definite integral}} \cr
& \int {\left( {2x + 1} \right){e^{ - x}}} dx = \left. {\left( { - \left( {2x + 1} \right){e^{ - x}} - 2{e^{ - x}}} \right)} \right|_0^1 \cr
& {\text{evaluating the limits}} \cr
& = \left( { - \left( {2\left( 1 \right) + 1} \right){e^{ - 1}} - 2{e^{ - 1}}} \right) - \left( { - \left( {2\left( 0 \right) + 1} \right){e^{ - 0}} - 2{e^{ - 0}}} \right) \cr
& {\text{simplifying}} \cr
& = \left( { - 3{e^{ - 1}} - 2{e^{ - 1}}} \right) - \left( { - 1 - 2} \right) \cr
& = - 5{e^{ - 1}} + 3 \cr
& = - \frac{5}{e} + 3 \cr} $$
