Answer
$\frac{3}{{\ln 3}} - \frac{2}{{{{\left( {\ln 3} \right)}^2}}}$
Work Step by Step
$$\eqalign{
& \int_0^1 {x \cdot {3^x}} dx \cr
& {\text{Integrate }}\int {x \cdot {3^x}} dx{\text{ by parts}} \cr
& {\text{Let }}u = x,{\text{ }}du = dx \cr
& dv = {3^x}dx,{\text{ }}v = \frac{{{3^x}}}{{\ln 3}} \cr
& {\text{Using the integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {x \cdot {3^x}} d\theta = x\left( {\frac{{{3^x}}}{{\ln 3}}} \right) - \int {\left( {\frac{{{3^x}}}{{\ln 3}}} \right)dx} \cr
& \int {x \cdot {3^x}} d\theta = \frac{{{3^x}}}{{\ln 3}}x - \frac{1}{{\ln 3}}\int {{3^x}dx} \cr
& {\text{Integrate}} \cr
& \int {x \cdot {3^x}} d\theta = \frac{{{3^x}}}{{\ln 3}}x - \frac{1}{{\ln 3}}\left( {\frac{{{3^x}}}{{\ln 3}}} \right) + C \cr
& \int {x \cdot {3^x}} d\theta = \frac{{{3^x}}}{{\ln 3}}x - \frac{{{3^x}}}{{{{\left( {\ln 3} \right)}^2}}} + C \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int_0^1 {x \cdot {3^x}} dx = \left[ {\frac{{{3^x}}}{{\ln 3}}x - \frac{{{3^x}}}{{{{\left( {\ln 3} \right)}^2}}}} \right]_0^1 \cr
& = \left[ {\frac{{{3^1}}}{{\ln 3}} - \frac{{{3^1}}}{{{{\left( {\ln 3} \right)}^2}}}} \right] - \left[ {\frac{{{3^0}}}{{\ln 3}}\left( 0 \right) - \frac{{{3^0}}}{{{{\left( {\ln 3} \right)}^2}}}} \right] \cr
& {\text{Simplify}} \cr
& = \frac{3}{{\ln 3}} - \frac{3}{{{{\left( {\ln 3} \right)}^2}}} + \frac{1}{{{{\left( {\ln 3} \right)}^2}}} \cr
& = \frac{3}{{\ln 3}} - \frac{2}{{{{\left( {\ln 3} \right)}^2}}} \cr} $$
