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Evaluate the indefinite integral of x2e-2x

by RoRi

Compute the following indefinite integral,

    \[ \int x^2 e^{-2x} \, dx. \]

To compute this integral we will integrate by parts twice. First, let

    \begin{align*} u &= x^2 & du &= 2x \, dx \\ dv &= e^{-2x} \, dx & v &= -\frac{1}{2} e^{-2x}. \end{align*}

Therefore we have

    \[ \int x^2 e^{-2x} \, dx = -\frac{x^2}{2} e^{-2x} + \int xe^{-2x} \, dx. \]

To evaluate this next integral we use integration by parts a second time with

    \begin{align*} u &= x & du &= dx \\ dv &= e^{-2x}\, dx & v &= -\frac{1}{2}e^{-2x}. \end{align*}

Giving us

    \begin{align*} \int xe^{-2x} \, dx &= -\frac{x}{2}e^{-2x} + \frac{1}{2} \int e^{-2x} \, dx \\ &= -\frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x} + C. \end{align*}

So, putting this back into our formula we have

    \begin{align*} \int x^2 e^{-2x} \, dx &= -\frac{x^2}{2} e^{-2x} + \int xe^{-2x} \, dx \\ &= -\frac{x^2}{2} e^{-2x} - \frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x} + C \\ &= -\frac{1}{2} e^{-2x} \left( x^2 + x + \frac{1}{2} \right) + C. \end{align*}

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