Answer
$ \int x e^{-2x} dx = \frac{-x e^{-2x}}{2} - \frac{e^{-2x}}{4} + C $
Work Step by Step
The given indefinite integral is ,
$\int x e^{-2x} dx$
Solve this with integration by parts below :
The formula of integration by parts is,
$ \int u dv = uv - \int v du$
Let's consider,
$ u = x$ and $ dv = e^{-2x} dx$
Therefore by taking the derivative of $u$, $ du = dx $ and by taking the anti-derivative of $dv$, $v = \frac{-1}{2} e^{-2x} $
So, according to integration by parts ,
$ \int x e^{-2x} dx = \frac{-1}{2} x e^{-2x} - \int \frac{-1}{2} e^{-2x} dx $
$ \int x e^{-2x} dx = \frac{- x e^{-2x}}{2} + \frac{1}{2} \int e^{-2x} dx $
$ \int x e^{-2x} dx = \frac{-1}{2} x e^{-2x} + \frac{1}{2} [ \frac{-1}{2} e^{-2x}] + C $
$ \int x e^{-2x} dx = \frac{-x e^{-2x}}{2} - \frac{e^{-2x}}{4} + C $
