AP Calculus BC

Product Rule vs Integration by Parts

Integration by parts is the product rule integrated and then rearranged, which is exactly where its minus sign comes from. The product rule differentiates a product outright and finishes.

Product rule

Use when: You are differentiating something written as one function times another, and you want the derivative directly.

Integration by parts

Use when: You are integrating a product whose two factors come from different families, such as a polynomial times an exponential, and no substitution or identity applies.

Side by side

	Product rule	Integration by parts
Direction	Differentiation	Integration
Statement	$(uv)' = u'v + uv'$	$\int u\,dv = uv - \int v\,du$
Sign between the terms	Plus, the two pieces add	Minus, created by moving a term to the other side
Treatment of the two factors	Symmetric, swapping them changes nothing	Asymmetric, one is differentiated and the other is integrated
What you end up holding	A finished derivative	A second integral, worth having when it is simpler, or when a second application reproduces the original integral so you can solve for it algebraically (as with $\int e^x \sin x\,dx$ )

Parts is not a new idea. Start from the product rule, integrate both sides with respect to $x$ , and solve for the piece you want. Two lines of algebra produce the whole formula, and every feature of it that looks arbitrary is a consequence of one of those lines.

\frac{d}{dx}(uv) = u'v + uv' \quad\Longrightarrow\quad uv = \int u'v\,dx + \int uv'\,dx \quad\Longrightarrow\quad \int uv'\,dx = uv - \int u'v\,dx

The derivation also assigns the labels. Whatever you call $u$ is the factor that gets differentiated, so pick the factor that improves under differentiation, and make sure the rest of the integrand is something you can antidifferentiate. In $\int x\,e^{x}\,dx$ , choosing $u = x$ sends $x$ to $1$ and leaves $\int e^{x}\,dx$ , giving $x e^{x} - e^{x} + C$ . Choosing $u = e^{x}$ instead turns $x$ into $\frac{x^2}{2}$ and leaves an integral worse than the original. A simpler second integral is not the only useful outcome, though: for $\int e^x \sin x\,dx$ parts returns $\int e^x \cos x\,dx$ , which is no easier at all, and a second application brings the original integral back, so you move it to the left side and solve to get $\frac{e^x(\sin x - \cos x)}{2} + C$ .

The mistake: carrying the product rule plus sign across

A plus sign carries over from differentiation, and students write $\int u\,dv = uv + \int v\,du$ out of habit. The minus exists only because $\int uv'\,dx$ was moved to one side of the equation. A parts answer built on a plus sign is wrong every time, and differentiating your result with the product rule catches it in one line.

Frequently asked questions

Is integration by parts just the product rule backwards?

It is the product rule integrated and rearranged, which is not quite the same as a clean reversal. The chain rule reverses into substitution and the problem is finished. Parts leaves you with $\int v\,du$ still to do, so it is a trade rather than an undoing.

How do I choose u in integration by parts?

Work down the list logarithmic, inverse trigonometric, algebraic, trigonometric, exponential, and take the first type present as $u$ . The point of that order is that logarithms and inverse trigonometric functions get much simpler when differentiated, while exponentials are just as easy to integrate as to differentiate.

Is integration by parts on the AP Calculus AB exam?

No. It is a BC-only topic in Unit 6. AB students are expected to handle products by substitution, by expanding, or by a trigonometric identity, and no AB question requires parts.

In the CED: Unit 2: Defining the Derivative, Unit 6: Integration and Accumulation