Answer
$$ - \frac{1}{8}{\left( {4 - 2x} \right)^4} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {4 - 2x} \right)}^3}} dx \cr
& {\text{substitute }}u = 4 - 2x,{\text{ }}du = - 2dx \cr
& \int {{{\left( {4 - 2x} \right)}^3}} dx = \int {{u^3}} \left( { - \frac{1}{2}du} \right) \cr
& = - \frac{1}{2}\int {{u^3}} du \cr
& {\text{power rule}} \cr
& - \frac{1}{2}\left( {\frac{{{u^4}}}{4}} \right) + C \cr
& = - \frac{1}{8}{u^4} + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{8}{\left( {4 - 2x} \right)^4} + C \cr} $$
