AP Calculus BC

Integral of x e^x: Answer, Parts Setup, Mistakes

Q: What is the integral of x e^x?

It is (x - 1)e^{x} + C, equivalently x e^{x} - e^{x} + C. Integration by parts with u = x and dv = e^{x}\,dx gives x e^{x} - \int e^{x}\,dx, and the leftover integral is just e^{x}.

Q: What is the definite integral of x e^x from 0 to 1?

It equals exactly 1. Evaluate (x - 1)e^{x} at the endpoints: at x = 1 the factor (x - 1) is zero, and at x = 0 the value is -1, so the difference is 0 - (-1) = 1.

Q: How do I integrate x^2 e^x or x e^(2x)?

Apply parts again for the first: \int x^{2}e^{x}\,dx = x^{2}e^{x} - 2\int x e^{x}\,dx = (x^{2} - 2x + 2)e^{x} + C. For the second, u = x and dv = e^{2x}\,dx gives v = \frac{e^{2x}}{2} and the answer e^{2x}\left(\frac{x}{2} - \frac{1}{4}\right) + C.

The integral of x e^x with respect to x is (x - 1)e^x + C. Integration by parts with u = x and dv = e^x dx gives x e^x minus the integral of e^x dx, so the answer is x e^x - e^x + C, which factors as (x - 1)e^x + C. Differentiating it back with the product rule returns x e^x.

\int x e^{x}\,dx = (x - 1)e^{x} + C

How to integrate x e^x by parts

The integrand $x e^{x}$ is a product of two different kinds of function, an algebraic factor and an exponential factor, and nothing inside it is the derivative of anything else in it. That rules out substitution and points at integration by parts, which comes from reversing the product rule.

\int u\,dv = uv - \int v\,du

Let $u$ be the factor that gets simpler when you differentiate it. Here $x$ differentiates to $1$ , while $e^{x}$ never simplifies, so take $u = x$ and let $e^{x}\,dx$ be $dv$ .

u = x, \quad du = dx, \quad dv = e^{x}\,dx, \quad v = e^{x}

Substitute those four pieces into the formula. The $uv$ term is $x e^{x}$ , and the new integral $\int v\,du$ is $\int e^{x}\,dx$ .

\int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx

The remaining integral is one of the basic antiderivatives, so a single pass through parts finishes the problem. Factor $e^{x}$ out of both terms to get the compact form.

\int x e^{x}\,dx = x e^{x} - e^{x} + C = (x - 1)e^{x} + C

Check it by differentiating

Differentiate $(x - 1)e^{x}$ with the product rule: $(1)e^{x} + (x - 1)e^{x} = e^{x} + x e^{x} - e^{x} = x e^{x}$ . The two $e^{x}$ terms cancel and you are back to the integrand, which is the fastest way to catch a sign slip on an exam.

Why u = x and not u = e^x

Integration by parts only helps if the new integral $\int v\,du$ is easier than the one you started with. Choosing $u = e^{x}$ and $dv = x\,dx$ is legal, but it makes $v = \frac{x^{2}}{2}$ and pushes the power up instead of down.

\int x e^{x}\,dx = \frac{x^{2}}{2}e^{x} - \int \frac{x^{2}}{2}e^{x}\,dx

The new integrand $\frac{x^{2}}{2}e^{x}$ is worse than $x e^{x}$ , and repeating the choice makes it worse again. The LIATE ordering is a shortcut for picking $u$ so this does not happen: take the first type on the list that appears in the integrand.

Logarithmic, such as $\ln x$
Inverse trigonometric, such as $\arctan x$
Algebraic, such as $x$ or $x^{2}$
Trigonometric, such as $\sin x$
Exponential, such as $e^{x}$

In $x e^{x}$ the algebraic factor $x$ outranks the exponential $e^{x}$ , so $u = x$ . The list is a heuristic, not a theorem, but the reason behind it is the real test: differentiating $u$ should shrink the problem, and antidifferentiating $dv$ should not grow it.

Where the integral of x e^x shows up on the AP exam

Integration by parts is Topic 6.11 in Unit 6 (Integration and Accumulation of Change), and it is BC only, so $\int x e^{x}\,dx$ will not be asked on AB. Unit 6 carries a weighting of 15 to 20 percent on both exams, and Topic 6.14 (Selecting Techniques for Antidifferentiation) is where you have to recognize that this integrand wants parts rather than substitution.

The definite version is the common exam form. With the antiderivative in factored form the arithmetic is short, because $(x - 1)e^{x}$ is $0$ at $x = 1$ .

\int_{0}^{1} x e^{x}\,dx = \left[(x - 1)e^{x}\right]_{0}^{1} = 0 - (-1) = 1

Two variants come up often enough to be worth recognizing. An exponent of $ax$ changes the bookkeeping but not the method, and a squared algebraic factor needs parts twice, dropping the power by one each pass.

\int x e^{ax}\,dx = e^{ax}\left(\frac{x}{a} - \frac{1}{a^{2}}\right) + C, \quad a \neq 0

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

That second result is the same machinery: one pass turns $\int x^{2}e^{x}\,dx$ into $x^{2}e^{x} - 2\int x e^{x}\,dx$ , and the integral you are left with is the one on this page.

Common mistakes with the integral of x e^x

Multiplying the antiderivatives. There is no product rule for integrals, so $\frac{x^{2}}{2}e^{x}$ is not the answer. Differentiate it and you get $x e^{x} + \frac{x^{2}}{2}e^{x}$ , not $x e^{x}$ .
Losing the minus sign in front of $\int v\,du$ and writing $x e^{x} + e^{x} + C$ . This is the single most common parts error, and one differentiation catches it.
Reaching for substitution. Setting $u = e^{x}$ gives $du = e^{x}\,dx$ , and there is no spare $e^{x}$ in $x e^{x}\,dx$ to absorb; setting $u = x$ leaves the $e^{x}$ untouched.
Writing $v = \frac{e^{x}}{x}$ or similar. Antidifferentiating $dv = e^{x}\,dx$ gives $v = e^{x}$ , because $e^{x}$ is its own antiderivative.
Factoring carelessly. From $x e^{x} - e^{x}$ the factored form is $(x - 1)e^{x}$ , not $(1 - x)e^{x}$ .

You do not need a second constant

When you antidifferentiate $dv$ to get $v$ , drop the constant. Carrying $v = e^{x} + k$ through the formula adds $kx$ and subtracts $kx$ , so it cancels every time. Write a single $+ C$ at the end of an indefinite integral, and none at all on a definite one.

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 e^x?

It is $(x - 1)e^{x} + C$ , equivalently $x e^{x} - e^{x} + C$ . Integration by parts with $u = x$ and $dv = e^{x}\,dx$ gives $x e^{x} - \int e^{x}\,dx$ , and the leftover integral is just $e^{x}$ .

Can I use u-substitution instead of integration by parts?

No. Substitution needs an inner function whose derivative is already sitting in the integrand up to a constant. In $x e^{x}\,dx$ , taking $u = e^{x}$ needs a spare $e^{x}$ that is not there, and taking $u = x$ changes nothing. The product of two unrelated factors is the signature of integration by parts.

Is the integral of x e^x on the AP Calculus AB exam?

No. Integration by parts is Topic 6.11, which the CED marks BC only, so this integral can appear on BC but not on AB. AB students still integrate $e^{x}$ and $e^{kx}$ by the basic rules and by substitution.

What is the definite integral of x e^x from 0 to 1?

It equals exactly $1$ . Evaluate $(x - 1)e^{x}$ at the endpoints: at $x = 1$ the factor $(x - 1)$ is zero, and at $x = 0$ the value is $-1$ , so the difference is $0 - (-1) = 1$ .

How do I integrate x^2 e^x or x e^(2x)?

Apply parts again for the first: $\int x^{2}e^{x}\,dx = x^{2}e^{x} - 2\int x e^{x}\,dx = (x^{2} - 2x + 2)e^{x} + C$ . For the second, $u = x$ and $dv = e^{2x}\,dx$ gives $v = \frac{e^{2x}}{2}$ and the answer $e^{2x}\left(\frac{x}{2} - \frac{1}{4}\right) + C$ .