Answer
$$\frac{{\left( {{x^2} - 1} \right){3^{ - x}}}}{2\ln 3} - \frac{{{3^{ - x + 1}}x}}{{{{\left( {\ln 3} \right)}^2}}} - \frac{{{3^{ - x + 1}}}}{{{{\left( {\ln 3} \right)}^3}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {{x^2} - 1} \right){3^{ - x}}} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = {x^2} - 1,{\text{ }}du = 2xdx{\text{ }} \cr
& dv = {3^{ - x}}dx,{\text{ }}v = \int {{3^{ - x}}dx} = - \frac{{{3^{ - x}}}}{{\ln 3}} \cr
& {\text{Using the formula of integration by parts}} \cr
& \int u dv = uv - \int v du \cr
& \int {\left( {{x^2} - 1} \right)} \left( {{3^{ - x}}} \right)dx = \left( {{x^2} - 1} \right)\left( { - \frac{{{3^{ - x}}}}{{\ln 3}}} \right) - \int {\left( { - \frac{{{3^{ - x}}}}{{\ln 3}}} \right)} \left( {2x} \right)dx \cr
& \int {\left( {{x^2} - 1} \right){3^{ - x}}} dx = \frac{{\left( {{x^2} - 1} \right){3^{ - x}}}}{2\ln 3} + \frac{3}{{\ln 3}}\int {x{3^{ - x}}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Integrate by parts }}\frac{3}{{\ln 3}}\int {x{3^{ - x}}} dx{\text{,}} \cr
& {\text{Let }}u = x,{\text{ }}du = dx{\text{ }} \cr
& dv = {3^{ - x}}dx,{\text{ }}v = \int {{3^{ - x}}dx} = - \frac{{{3^{ - x}}}}{{\ln 3}} \cr
& \int u dv = uv - \int v du \cr
& \frac{3}{{\ln 3}}\int {x{3^{ - x}}} dx = \frac{3}{{\ln 3}}\left( x \right)\left( { - \frac{{{3^{ - x}}}}{{\ln 3}}} \right) - \frac{3}{{\ln 3}}\int {\left( { - \frac{{{3^{ - x}}}}{{\ln 3}}} \right)dx} \cr
& = - \frac{{{3^{ - x + 1}}x}}{{{{\left( {\ln 3} \right)}^2}}} + \frac{3}{{{{\left( {\ln 3} \right)}^2}}}\int {{3^{ - x}}dx} \cr
& = - \frac{{{3^{ - x + 1}}x}}{{{{\left( {\ln 3} \right)}^2}}} + \frac{3}{{{{\left( {\ln 3} \right)}^2}}}\left( { - \frac{{{3^{ - x}}}}{{\ln 3}}} \right) + C \cr
& = - \frac{{{3^{ - x + 1}}x}}{{{{\left( {\ln 3} \right)}^2}}} - \frac{{{3^{ - x + 1}}}}{{{{\left( {\ln 3} \right)}^3}}} + C \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& \int {\left( {{x^2} - 1} \right){3^{ - x}}} dx = \frac{{\left( {{x^2} - 1} \right){3^{ - x}}}}{2\ln 3} - \frac{{{3^{ - x + 1}}x}}{{{{\left( {\ln 3} \right)}^2}}} - \frac{{{3^{ - x + 1}}}}{{{{\left( {\ln 3} \right)}^3}}} + C \cr} $$
