We are asked to evaluate the indefinite integral $\int xe^{-2x} dx$.

1. Problem Description

We are asked to evaluate the indefinite integral $\int xe^{-2x} dx$.

2. Solution Steps

We will use integration by parts to evaluate the integral. The formula for integration by parts is

$\int u dv = uv - \int v du$.

Let $u = x$ and $dv = e^{-2x} dx$. Then $du = dx$ and $v = \int e^{-2x} dx = -\frac{1}{2}e^{-2x}$.

Substituting these values into the integration by parts formula, we have

$\int xe^{-2x} dx = x(-\frac{1}{2}e^{-2x}) - \int (-\frac{1}{2}e^{-2x}) dx$

$= -\frac{1}{2}xe^{-2x} + \frac{1}{2}\int e^{-2x} dx$

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

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

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

3. Final Answer

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