Question

Integralplease help me

Answer 1

Recall that the product rule for derivatives tells us that, for two functions $f=f(x)$ and $g=g(x)$,

$\dfrac{\mathrm d}{\mathrm dx}[fg]=\dfrac{\mathrm df}{\mathrm dx}g+f\dfrac{\mathrm dg}{\mathrm dx}$

Integrating both sides with respect to $x$ gives the reverse "rule" for integration:

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}[fg]\,\mathrm dx=\int\frac{\mathrm df}{\mathrm dx}g\,\mathrm dx+\int f\dfrac{\mathrm dg}{\mathrm dx}\,\mathrm dx$
$\implies \displaystyle\int f\frac{\mathrm dg}{\mathrm dx}\,\mathrm dx=fg-\int\dfrac{\mathrm df}{\mathrm dx}g\,\mathrm dx$

You might know this process by the name "integration by parts". This is the standard method for the given integral.

Take

$f=x\implies\dfrac{\mathrm df}{\mathrm dx}=1$
$\dfrac{\mathrm dg}{\mathrm dx}=e^{-2x}\implies g=-\dfrac12e^{-2x}$

and so

$\displaystyle\int xe^{-2x}\,\mathrm dx=-\dfrac x2e^{-2x}+\dfrac12\int e^{-2x}\,\mathrm dx$

The remaining integral is trivial, and the overall result is

$\displaystyle\int xe^{-2x}\,\mathrm dx=-\dfrac x2e^{-2x}-\dfrac14e^{-2x}+C$