Answer
$$ - \frac{1}{{30}}{\left( {1 - 3{x^2}} \right)^5} + C$$
Work Step by Step
$$\eqalign{
& \int {x{{\left( {1 - 3{x^2}} \right)}^4}} dx \cr
& {\text{Integrate by substitution}} \cr
& u = 1 - 3{x^2},{\text{ }}du = - 6xdx,{\text{ }}dx = - \frac{1}{{6x}}du \cr
& {\text{Substituting}} \cr
& \int {x{{\left( {1 - 3{x^2}} \right)}^4}} dx = \int {x{u^4}} \left( { - \frac{1}{{6x}}} \right)du \cr
& = \int {{u^4}} \left( { - \frac{1}{6}} \right)du \cr
& = - \frac{1}{6}\int {{u^4}} du \cr
& = - \frac{1}{6}\left( {\frac{{{u^5}}}{5}} \right) + C \cr
& = - \frac{1}{{30}}{u^5} + C \cr
& {\text{Write in terms of }}x \cr
& = - \frac{1}{{30}}{\left( {1 - 3{x^2}} \right)^5} + C \cr} $$
