AP Calculus BC

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

Q: Why choose u = x^2 and not dv = x^2 dx?

Because differentiating x^2 drives its power down toward a constant, which ends the process, while e^x integrates to itself and never gets worse. Letting dv = x^2\,dx would raise the power and make the integral harder.

Q: How many times do I apply integration by parts?

Twice, matching the exponent on x. Each pass lowers x^2 \to 2x \to constant. In general \int x^n e^x\,dx needs n applications.

The integral of x^2 e^x is e^x(x^2 - 2x + 2) + C. You get it by integration by parts done twice: each pass lowers the power of x by one, from x^2 to 2x to a constant, until the exponential integrates on its own.

\int x^2 e^x\,dx = e^x(x^2-2x+2) + C

First pass of integration by parts

Integration by parts, $\int u\,dv = uv - \int v\,du$ , trades a hard integral for an easier one. Differentiating the polynomial is what makes progress, so let $u = x^2$ and $dv = e^x\,dx$ .

u = x^2,\quad dv = e^x\,dx \Rightarrow du = 2x\,dx,\quad v = e^x

That leaves a new integral with $x$ only to the first power.

\int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx

Second pass and the result

Apply parts again to $\int 2x e^x\,dx$ with $u = 2x$ and $dv = e^x\,dx$ , so $du = 2\,dx$ and $v = e^x$ .

\int 2x e^x\,dx = 2x e^x - \int 2e^x\,dx = 2x e^x - 2e^x

Substitute back and factor out $e^x$ .

\int x^2 e^x\,dx = x^2 e^x - (2x e^x - 2e^x) = e^x(x^2 - 2x + 2) + C

Differentiating $e^x(x^2-2x+2)$ by the product rule returns $e^x(x^2-2x+2) + e^x(2x-2) = x^2 e^x$ , confirming the answer.

The mistake students make

The frequent slip is a sign error on the second pass, dropping the minus in front of $\int 2x e^x\,dx$ and writing $x^2 e^x + 2x e^x - 2e^x$ . Track the leading minus carefully, since both terms inside the parentheses change sign when the bracket is expanded.

Alternating signs, one power at a time

For $x^n e^x$ , parts applied $n$ times gives $e^x$ times a polynomial whose signs alternate, here $x^2 - 2x + 2$ . A tabular (DI) layout keeps those alternating signs organized.

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 choose u = x^2 and not dv = x^2 dx?

Because differentiating $x^2$ drives its power down toward a constant, which ends the process, while $e^x$ integrates to itself and never gets worse. Letting $dv = x^2\,dx$ would raise the power and make the integral harder.

How many times do I apply integration by parts?

Twice, matching the exponent on $x$ . Each pass lowers $x^2 \to 2x \to$ constant. In general $\int x^n e^x\,dx$ needs $n$ applications.

Is there a faster way than writing out both passes?

Yes, tabular integration (the DI method) lists derivatives of $x^2$ beside integrals of $e^x$ and reads off the answer with alternating signs. It gives the same $e^x(x^2 - 2x + 2)$ with less writing.