AP Calculus AB and BC

Integral of x^2 e^(x^3): Substitution

Q: What about e^(x^3) on its own?

It has no elementary antiderivative. The x^{2} factor is what makes this version solvable.

The integral of x squared times e to the x cubed is e to the x cubed over 3, plus C. The substitution u equals x cubed needs 3x squared dx, and the x squared already present supplies it up to the factor of 3.

\int x^{2}e^{x^{3}}\,dx = \frac{e^{x^{3}}}{3} + C

The spare factor is the whole point

u = x^{3}, \quad du = 3x^{2}dx \implies x^{2}dx = \frac{du}{3}

\int x^{2}e^{x^{3}}dx = \frac{1}{3}\int e^{u}du = \frac{e^{x^{3}}}{3} + C

Remove the x^2 and it becomes impossible

The integral of e^(x^3) alone has no elementary antiderivative. The x^2 in front is not decoration; it is exactly what the substitution consumes.

How to spot it

Look for an inner function whose derivative is already present up to a constant. Here the inside is $x^{3}$ and its derivative $3x^{2}$ matches the $x^{2}$ outside. That check is Topic 6.9 in one sentence.

Common mistakes

Answering $\frac{x^{3}e^{x^{3}}}{3}$ , treating $e^{x^{3}}$ as if the power rule applied to it.
Forgetting the $\frac{1}{3}$ .
Trying integration by parts. Substitution finishes it in one line.

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

What is the integral of x^2 e^(x^3)?

It is $\frac{e^{x^{3}}}{3} + C$ .

What about e^(x^3) on its own?

It has no elementary antiderivative. The $x^{2}$ factor is what makes this version solvable.