Evaluate the indefinite integral of x3e-x2

Compute the following indefinite integral,

$\int x^3 e^{-x^2} \, dx.$

First, we rearrange the integrand a bit to get a form in which we can make a $u$ -substitution,

$\int x^3 e^{-x^2} \, dx = -\frac{1}{2} \int x^2(-2xe^{-x^2} ) \, dx.$

Then, let

$u = e^{-x^2}, \quad du = -2x e^{-x^2} \, dx \quad \implies \quad \log u = -x^2.$

Therefore, making a $u$ -substitution, we have

$\begin{align*} \int x^3 e^{-x^2} \, dx &= -\frac{1}{2} \int x^2 (-2xe^{-x^2} ) \, dx \\[10pt] &= -\frac{1}{2} \int -\log u \, du \\[10pt] &= \frac{1}{2} \int \log u \, du \\[10pt] &= \frac{1}{2} (u \log u - u) + C \\[10pt] &= \frac{1}{2} \left( e^{-x^2} \log \left( e^{-x^2} \right) - e^{-x^2} \right) + C \\[10pt] &= -\frac{1}{2} (x^2+1)e^{-x^2} + C. \end{align*}$

Stumbling Robot

Evaluate the indefinite integral of x³e^-x²

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