Answer
$$ - \frac{1}{{2\ln 4}}\left( {{4^{ - {x^2}}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {x{4^{ - {x^2}}}} dx \cr
& {\text{substitute }}u = - {x^2} \cr
& du = - 2xdx \cr
& - \frac{1}{2}du = xdx \cr
& \int {x{4^{ - {x^2}}}} dx \cr
& = \int {{4^u}\left( { - \frac{1}{2}du} \right)} \cr
& = - \frac{1}{2}\int {{4^u}} du \cr
& {\text{find the antiderivative}} \cr
& = - \frac{1}{2}\left( {\frac{{{4^u}}}{{\ln 4}}} \right) + C \cr
& = - \frac{1}{{2\ln 4}}\left( {{4^u}} \right) + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = - {x^2} \cr
& = - \frac{1}{{2\ln 4}}\left( {{4^{ - {x^2}}}} \right) + C \cr} $$
