AP Calculus BC

Integral of x^3 e^(x^2): Substitute, Then Parts

Q: What is the antiderivative of x^3 e^(x^2)?

It works out to \frac{e^{x^{2}}\left(x^{2}-1\right)}{2} + C.

Q: Can integration by parts do this without substituting first?

Yes, if you split the integrand well. Taking dv = e^{x^{2}}dx alone fails, since e^{x^{2}} has no elementary antiderivative, but u = x^{2} with dv = x e^{x^{2}}dx gives the elementary v = \frac{e^{x^{2}}}{2}. Parts then finishes in one step: \frac{x^{2}e^{x^{2}}}{2} - \int x e^{x^{2}}dx = \frac{e^{x^{2}}\left(x^{2}-1\right)}{2} + C. Substituting first simply saves you from having to spot that v.

Substituting u equal to x squared leaves one half the integral of u e to the u, and parts finishes that in one step. The result is e to the x squared times x squared minus 1, all over 2, plus C.

\int x^{3}e^{x^{2}}\,dx = \frac{e^{x^{2}}\left(x^{2}-1\right)}{2} + C

Substitute first

Split the cubic as $x^{3} = x^{2}\cdot x$ . With $u = x^{2}$ you get $du = 2x\,dx$ , so $x\,dx = \frac{1}{2}du$ , and the spare $x^{2}$ becomes $u$ .

\int x^{3}e^{x^{2}}\,dx = \int x^{2}e^{x^{2}}\cdot x\,dx = \frac{1}{2}\int u\,e^{u}\,du

Every $x$ is gone, which is the test that a substitution has actually worked.

Then parts

The integral $\int u\,e^{u}\,du$ is the standard parts problem, the same one as $\int x e^{x}\,dx$ : differentiate the polynomial factor, integrate the exponential factor, and the polynomial disappears in one step.

\frac{1}{2}\int u\,e^{u}\,du = \frac{1}{2}\left(u e^{u} - e^{u}\right) = \frac{e^{u}\left(u-1\right)}{2}

\int x^{3}e^{x^{2}}\,dx = \frac{e^{x^{2}}\left(x^{2}-1\right)}{2} + C

Order of operations

Taking dv = e^(x^2) dx on its own gets you nowhere, because e^(x^2) has no antiderivative in elementary functions. Bundle the spare x into dv instead and parts works directly: u = x^2 with dv = x e^(x^2) dx gives v = e^(x^2)/2, and one step lands on the same answer. Substituting first is the route that does not depend on spotting that v.

The mistakes students make

The failures here are about strategy and constants rather than about the parts formula itself.

Taking $dv = e^{x^{2}}dx$ and inventing $v = \frac{e^{x^{2}}}{2x}$ . Differentiating that by the quotient rule does not return $e^{x^{2}}$ , and no elementary $v$ exists.
Losing the $\frac{1}{2}$ from $du = 2x\,dx$ and answering $e^{x^{2}}\left(x^{2}-1\right)$ , whose derivative is $2x^{3}e^{x^{2}}$ , double the integrand.
Substituting $u = x^{2}$ but leaving a stray $x$ , as in $\frac{1}{2}\int x\,u\,e^{u}\,du$ . A mixed integrand cannot be integrated in either variable.

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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 antiderivative of x^3 e^(x^2)?

It works out to $\frac{e^{x^{2}}\left(x^{2}-1\right)}{2} + C$ .

Can integration by parts do this without substituting first?

Yes, if you split the integrand well. Taking $dv = e^{x^{2}}dx$ alone fails, since $e^{x^{2}}$ has no elementary antiderivative, but $u = x^{2}$ with $dv = x e^{x^{2}}dx$ gives the elementary $v = \frac{e^{x^{2}}}{2}$ . Parts then finishes in one step: $\frac{x^{2}e^{x^{2}}}{2} - \int x e^{x^{2}}dx = \frac{e^{x^{2}}\left(x^{2}-1\right)}{2} + C$ . Substituting first simply saves you from having to spot that $v$ .

How do I know to substitute before using parts?

Check whether an inner function has its derivative present, up to a constant. Here $x^{2}$ has derivative $2x$ and there are odd powers of $x$ available, so substitution is the safe opening move and parts cleans up what is left.