Answer
$$ - \left( {x + 2} \right){e^{ - 3x}} - \frac{1}{3}{e^{ - 3x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {3x + 6} \right){e^{ - 3x}}} dx \cr
& {\text{setting }}\,\,\,\,\,\,\,u = 3x + 6{\text{ then }}du = 3dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{ - 3x}}dx{\text{ then }}v = - \frac{1}{3}{e^{ - 3x}} \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\left( {3x + 6} \right){e^{ - 3x}}} dx = \left( {3x + 6} \right)\left( { - \frac{1}{3}{e^{ - 3x}}} \right) - \int {\left( { - \frac{1}{3}{e^{ - 3x}}} \right)\left( {3dx} \right)} \cr
& {\text{simplifying}} \cr
& = 3\left( {x + 2} \right)\left( { - \frac{1}{3}{e^{ - 3x}}} \right) + \int {{e^{ - 3x}}dx} \cr
& = - \left( {x + 2} \right){e^{ - 3x}} + \int {{e^{ - 3x}}dx} \cr
& {\text{integrate using }}\int {{e^{kx}}dx} = \frac{{{e^{kx}}}}{k} + C \cr
& = - \left( {x + 2} \right){e^{ - 3x}} + \left( {\frac{{{e^{ - 3x}}}}{{ - 3}}} \right) + C \cr
& = - \left( {x + 2} \right){e^{ - 3x}} - \frac{1}{3}{e^{ - 3x}} + C \cr} $$
