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

Step-by-step solution

1. 1

   Notice why this one is doable at all. The closely related $\int e^{-x^2}dx$ has no elementary antiderivative. The extra factor of $x$ changes everything: it is (up to a constant) the derivative of the exponent, so the integral is a pure substitution. Recognising that distinction is the entire insight.

2. 2

   Choose the substitution. Let

   $u=-x^{2}\quad\Longrightarrow\quad du=-2x\,dx\quad\Longrightarrow\quad x\,dx=-\tfrac12\,du$

3. 3

   Rewrite the integral entirely in $u$. The $x\,dx$ in the original is replaced wholesale, leaving nothing in $x$ behind:

   $\int x\,e^{-x^{2}}\,dx=\int e^{u}\left(-\tfrac12\,du\right)=-\tfrac12\int e^{u}\,du$

4. 4

   Integrate and substitute back.

   $-\tfrac12 e^{u}+C=-\tfrac12 e^{-x^{2}}+C$

5. 5

   Differentiate to check. By the chain rule,

   $\frac{d}{dx}\left(-\tfrac12 e^{-x^{2}}\right)=-\tfrac12 e^{-x^{2}}\cdot(-2x)=x\,e^{-x^{2}}$

   which is the original integrand ✓. A numerical spot check at $x=1.3$: the symmetric difference quotient of $-\tfrac12 e^{-x^2}$ gives $0.2398754$, and $x e^{-x^2}=0.2398754$ ✓.

6. 6

   Note what a definite version would give. With limits $a$ to $b$ the answer is $\tfrac12\left(e^{-a^2}-e^{-b^2}\right)$; in particular $\int_0^{\infty}x e^{-x^2}dx=\tfrac12$.