AP Calculus AB and BC

Integral of x e^(x^2): Answer, Proof, and Steps

Q: Why can you integrate x e^(x^2) but not e^(x^2)?

Because the extra factor x matches the derivative of the exponent x^2, which is 2x. That match lets u = x^2 absorb the x\,dx. Plain e^{x^2} has no such factor and no elementary antiderivative.

Q: What u do you choose for x e^(x^2)?

Let u = x^2, the exponent. Then du = 2x\,dx, so x\,dx = \frac{du}{2}, and the integral becomes \frac{1}{2}\int e^u\,du = \frac{e^{x^2}}{2} + C.

The integral of x e^(x^2) is e^(x^2)/2 + C. Substituting u = x^2 gives du = 2x dx, so x dx = du/2 and the integral becomes one half times the integral of e^u. Differentiating e^(x^2)/2 gives (1/2) times 2x times e^(x^2), which is x e^(x^2).

\int xe^{x^2}\,dx = \frac{e^{x^2}}{2} + C

The substitution, step by step

The integrand pairs $e^{x^2}$ with a factor of $x$ . Since the derivative of the inner function $x^2$ is $2x$ , that stray $x$ is exactly what a u-substitution needs.

u = x^2 \quad\Longrightarrow\quad \frac{du}{dx} = 2x \quad\Longrightarrow\quad x\,dx = \frac{du}{2}

Replace $x^2$ with $u$ and $x\,dx$ with $\frac{du}{2}$ , then pull the constant outside.

\int xe^{x^2}\,dx = \int e^{u}\cdot\frac{du}{2} = \frac{1}{2}\int e^{u}\,du = \frac{1}{2}e^{u} + C

Substituting $u = x^2$ back gives the answer in the original variable.

\int xe^{x^2}\,dx = \frac{e^{x^2}}{2} + C

The x is not optional

Substitution works here only because the factor $x$ matches the derivative of $x^2$ up to the constant $2$ . Without that $x$ , the integral $\int e^{x^2}\,dx$ has no elementary antiderivative.

Checking by differentiating

Differentiate the result with the chain rule and confirm the $x$ returns.

\frac{d}{dx}\left(\frac{e^{x^2}}{2}\right) = \frac{1}{2}\cdot e^{x^2}\cdot 2x = xe^{x^2}

The chain rule brings out the inner derivative $2x$ , and that $2$ cancels the $\frac{1}{2}$ , leaving $xe^{x^2}$ exactly.

Common mistakes

Forgetting the $\frac{1}{2}$ . Without it the derivative is $2xe^{x^2}$ , twice the integrand.
Trying the power rule on the exponent. The variable sits in the exponent, so there is no power of $x$ to raise.
Attempting to integrate $e^{x^2}$ on its own. That has no elementary antiderivative; the outer $x$ is what makes this integral doable.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

Why can you integrate x e^(x^2) but not e^(x^2)?

Because the extra factor $x$ matches the derivative of the exponent $x^2$ , which is $2x$ . That match lets $u = x^2$ absorb the $x\,dx$ . Plain $e^{x^2}$ has no such factor and no elementary antiderivative.

What is the integral of x e^(x^2) from 0 to 1?

$\int_0^1 xe^{x^2}\,dx = \left[\frac{e^{x^2}}{2}\right]_0^1 = \frac{e - 1}{2}$ .

What u do you choose for x e^(x^2)?

Let $u = x^2$ , the exponent. Then $du = 2x\,dx$ , so $x\,dx = \frac{du}{2}$ , and the integral becomes $\frac{1}{2}\int e^u\,du = \frac{e^{x^2}}{2} + C$ .