Question $4$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}x{e}^{-2x}dx=$

$-2x{e}^{-2x}+\frac{1}{2}{e}^{-2x}+C$

$\frac{1}{2}{e}^{-2x}(4x+1)+C$

$\frac{-1}{2}x{e}^{-2x}-\frac{1}{4}{e}^{-2x}+C$

$\frac{1}{4}{e}^{-2x}(2x-1)+C$

None $of$ these