Recall that the product rule for derivatives tells us that, for two functions [tex]f=f(x)[/tex] and [tex]g=g(x)[/tex],
[tex]\dfrac{\mathrm d}{\mathrm dx}[fg]=\dfrac{\mathrm df}{\mathrm dx}g+f\dfrac{\mathrm dg}{\mathrm dx}[/tex]
Integrating both sides with respect to [tex]x[/tex] gives the reverse "rule" for integration:
[tex]\displaystyle\int\frac{\mathrm d}{\mathrm dx}[fg]\,\mathrm dx=\int\frac{\mathrm df}{\mathrm dx}g\,\mathrm dx+\int f\dfrac{\mathrm dg}{\mathrm dx}\,\mathrm dx[/tex]
[tex]\implies \displaystyle\int f\frac{\mathrm dg}{\mathrm dx}\,\mathrm dx=fg-\int\dfrac{\mathrm df}{\mathrm dx}g\,\mathrm dx[/tex]
You might know this process by the name "integration by parts". This is the standard method for the given integral.
Take
[tex]f=x\implies\dfrac{\mathrm df}{\mathrm dx}=1[/tex]
[tex]\dfrac{\mathrm dg}{\mathrm dx}=e^{-2x}\implies g=-\dfrac12e^{-2x}[/tex]
and so
[tex]\displaystyle\int xe^{-2x}\,\mathrm dx=-\dfrac x2e^{-2x}+\dfrac12\int e^{-2x}\,\mathrm dx[/tex]
The remaining integral is trivial, and the overall result is
[tex]\displaystyle\int xe^{-2x}\,\mathrm dx=-\dfrac x2e^{-2x}-\dfrac14e^{-2x}+C[/tex]
