Find the indefinite integral of x^2 * e^(-x^2) with respect to x.

Step-by-step solution

1. 1

   Split $x^{2}$ so that one $x$ joins the exponential. The factor $xe^{-x^{2}}$ has an elementary antiderivative (its inner derivative is $-2x$), while $e^{-x^{2}}$ alone does not. So set

   $u=x,\qquad dv=xe^{-x^{2}}\,dx.$

   Splitting the other way, with $dv=e^{-x^{2}}dx$, immediately stalls — the choice of pairing is the entire trick.

2. 2

   Find $v$ by substitution. With $t=-x^{2}$, $dt=-2x\,dx$:

   $v=\int xe^{-x^{2}}dx=-\frac12\int e^{t}\,dt=-\frac12 e^{-x^{2}}.$

   Also $du=dx$. The $-\tfrac12$ comes from $\frac{1}{-2}$ in the substitution and must be carried through.

3. 3

   Apply integration by parts. Using $\int u\,dv=uv-\int v\,du$:

   $\int x^{2}e^{-x^{2}}dx=x\left(-\frac12 e^{-x^{2}}\right)-\int\left(-\frac12 e^{-x^{2}}\right)dx=-\frac{x}{2}e^{-x^{2}}+\frac12\int e^{-x^{2}}dx.$

   The remaining integral is the Gaussian — simpler than what we started with, which is the sign that the pairing was chosen well.

4. 4

   Express the Gaussian with the error function. $\int e^{-x^{2}}dx$ has no elementary antiderivative. The error function is defined precisely to name it:

   $\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}dt\;\Longrightarrow\;\int e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}\operatorname{erf}(x)+C.$

5. 5

   Assemble and verify. Substituting back,

   $\int x^{2}e^{-x^{2}}dx=-\frac{x}{2}e^{-x^{2}}+\frac{\sqrt{\pi}}{4}\operatorname{erf}(x)+C.$

   Differentiating: $-\tfrac12 e^{-x^{2}}+x^{2}e^{-x^{2}}+\tfrac{\sqrt{\pi}}{4}\cdot\tfrac{2}{\sqrt{\pi}}e^{-x^{2}}=x^{2}e^{-x^{2}}$ ✓ — the two half-terms cancel exactly. A numeric check at $x=0.9$ gives $0.3603350$ both ways ✓. As a bonus, since $\operatorname{erf}(\infty)=1$ and the first term vanishes, $\int_{0}^{\infty}x^{2}e^{-x^{2}}dx=\tfrac{\sqrt{\pi}}{4}$.