(5) \( \int x e^{-2 x} d x \)

Calculate the integral \( \int x e^{-2x} dx \). Evaluate the integral by following steps: - step0: Evaluate using partial integration formula: \(\int xe^{-2x} dx\) - step1: Prepare for integration by parts: \(\begin{align}&u=x\\&dv=e^{-2x}dx\end{align}\) - step2: Calculate the derivative: \(\begin{align}&du=dx\\&dv=e^{-2x}dx\end{align}\) - step3: Evaluate the integral: \(\begin{align}&du=dx\\&v=-\frac{1}{2e^{2x}}\end{align}\) - step4: Substitute the values into formula: \(x\left(-\frac{1}{2e^{2x}}\right)-\int 1\times \left(-\frac{1}{2e^{2x}}\right) dx\) - step5: Calculate: \(-\frac{x}{2e^{2x}}-\int -\frac{1}{2e^{2x}} dx\) - step6: Use properties of integrals: \(-\frac{x}{2e^{2x}}+\int \frac{1}{2e^{2x}} dx\) - step7: Evaluate the integral: \(-\frac{x}{2e^{2x}}-\frac{1}{4e^{2x}}\) - step8: Add the constant of integral C: \(-\frac{x}{2e^{2x}}-\frac{1}{4e^{2x}} + C, C \in \mathbb{R}\) The integral of \(x e^{-2x}\) with respect to \(x\) is given by: \[ \int x e^{-2x} dx = -\frac{x}{2e^{2x}} - \frac{1}{4e^{2x}} + C \] where \(C\) is the constant of integration.