Integration by parts is the second major antidifferentiation technique AP Calculus BC candidates must master, sitting just below u-substitution in the syllabus. The rule itself is a one-line identity, but the way it is tested on the AP exam rewards candidates who can choose the right u and dv, recognise when a second application is required, and stay calm when the integral comes back with a piece of the original still sitting inside. For readers preparing through the YÖS or TR-YÖS pathway, the technique is doubly useful: the same machinery shows up in the Turkish university entrance mathematics component, and the structural habit of breaking a product into u·dv trains a discipline that carries across the rest of the BC syllabus. This article walks through the formula, the LIATE selection rule, the BC-only question families, the tabulation method, and the common pitfalls — all framed for a candidate who wants a clean, defensible plan for the integration portion of the AP Calculus BC exam.
The integration-by-parts formula and where it comes from
The technique is a direct consequence of the product rule for differentiation. Start with the product rule in its standard form, integrate both sides, and rearrange:
d/dx [u(x)·v(x)] = u'(x)·v(x) + u(x)·v'(x)
Integrating both sides with respect to x gives u·v = ∫u'·v dx + ∫u·v' dx. Solving for the integral of u·v' produces the form most textbooks write:
∫u dv = u·v − ∫v du
That is the entire identity. What separates a strong AP Calculus BC candidate from a struggling one is not the formula itself — it is the choice of u and dv, and the recognition that the technique converts a hard product integral into a second integral that is supposed to be easier. If the second integral is harder than the first, the choice was wrong, and the candidate should restart with a different split.
On the AP exam, integration by parts is most often the second step in a multi-step problem. The first step is usually u-substitution on one factor of the integrand, or a trigonometric rewrite. Candidates who try to apply integration by parts to the raw integrand without first simplifying waste two to three minutes and frequently arrive at a mess they cannot unwind. A defensible habit is to spend 15 to 30 seconds on inspection: classify the integrand as a product, identify the polynomial, logarithmic, inverse-trig, algebraic, and trigonometric factors, and only then commit to a choice.
For YÖS and TR-YÖS candidates, the same identity appears in a slightly different costume. The Turkish university entrance mathematics component tests the integration of products such as x·cos x, x·ex, and lnx, which are exactly the canonical AP-style problems. Practising the AP form therefore doubles as YÖS practice, and vice versa. A YÖS candidate who can do these integrals on a 90-second budget inside the timed AP section has a clear edge on the Turkish exam's slower, written-response format.
One final note on the formula: the indefinite integral on the right-hand side must include the constant of integration. AP graders are forgiving on missing +C in intermediate steps, but they will mark it in the final answer. Train yourself to write +C only once, at the very end of the whole computation, and never after the intermediate ∫v du term.
When to suspect the technique is the right tool
Three structural clues point to integration by parts. The integrand is the product of two functions from different families — for example, an algebraic factor times an exponential or trigonometric factor, or a logarithmic factor times anything. A substitution has been attempted and the result is still a product, with no new variable cleanly emerging. The derivative of one factor visibly cancels into the other. If any of these is true, the next move is the LIATE selection rule.
The LIATE selection rule and how to apply it
LIATE is a mnemonic, not a theorem, and AP readers do not award points for citing it. What it gives the candidate is a defensible default when the choice of u is not obvious. The letters stand, in order of preference, for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. The factor in the higher position in the list is typically chosen as u, and the remaining factor is paired with dx to form dv. The reason this works in practice is that differentiating a logarithm or inverse-trig function produces a factor that simplifies the integral, while integrating an exponential or trigonometric function gives another expression of the same family that is easier to handle.
Consider the canonical example ∫x·ex dx. L, I, A, T, E. The algebraic factor is in the A position; the exponential is in the E position. Since A comes before E in the priority list, u = x and dv = ex dx. The integration then proceeds with du = dx and v = ex, giving x·ex − ∫ex dx = x·ex − ex + C. The second integral was easier than the first, which is the sanity check the technique is functioning correctly.
Now reverse the integrand: ∫ex·x dx is the same product, but if the candidate applies LIATE correctly the answer is unchanged. If the candidate reverses the choice — taking u = ex and dv = x dx — the second integral becomes ∫x·ex dx, which is the original integral. The technique fails to simplify, and the candidate must restart. AP multiple-choice questions sometimes include both the LIATE-correct and LIATE-incorrect answers as distractors; the trap is real.
For YÖS and TR-YÖS preparation, LIATE has an additional benefit: the Turkish exam tends to test the same five or six integrand families, and the mnemonic generalises cleanly to those problems without modification. A candidate who has internalised the priority order does not have to re-derive the choice on every problem; the habit is portable.
Limitations of the mnemonic
LIATE is a heuristic, and a small number of BC problems will not be served by it. The classic case is ∫x2·sin x dx. LIATE recommends u = x2 because the algebraic factor is the priority, but applying the technique once reduces to ∫sin x dx inside an outer x2 coefficient, and the second integral is still a product. The technique must be applied twice, or the candidate should switch to the tabulation method covered later in this article.
The reduction formula for ∫xn·sin(ax) and similar families
BC-only questions sometimes ask the candidate to evaluate an integral such as ∫x2·ex dx or ∫x3·cos x dx, where the algebraic factor has degree two or higher. A single application of integration by parts is not enough; the technique must be applied repeatedly, with the new u chosen as the next power of x at each step. The pattern that emerges across these problems is called a reduction formula, and recognising it on sight is one of the strongest signals of BC-level command.
Take ∫x2·ex dx. First application: u = x2, dv = ex dx, giving x2·ex − ∫2x·ex dx. The new integral is still of the same form but with degree one in x. Second application: u = 2x, dv = ex dx, giving 2x·ex − ∫2·ex dx = 2x·ex − 2ex. Substituting back: x2·ex − [2x·ex − 2ex] = x2·ex − 2x·ex + 2ex + C.
The pattern generalises. For ∫xn·ex dx, each application peels off one power of x and adds another n·ex term to the chain, until the leftover integral is a constant times ex. The final answer always has the form (polynomial of degree n)·ex + C. AP candidates who recognise this signature can confirm their work by inspection at the end of a problem, and the recognition is testable across both the multiple-choice and free-response sections.
For trigonometric families such as ∫xn·cos x dx and ∫xn·sin x dx, the same reduction logic applies, but with a twist: the second integral will alternate between sine and cosine. After two applications, the original integral reappears on the right-hand side, but with a sign change. The technique is to move that reappearance to the left side of the equation and solve for the original integral algebraically. This is one of the few places in the BC syllabus where the candidate must rearrange a formula mid-problem, and the algebra itself is often worth a point on the free-response section.
Worked pattern: ∫ex·sin x dx
Apply integration by parts twice with the same choices: u = sin x, dv = ex dx on the first pass, then u = cos x, dv = ex dx on the second. The result is I = sin x·ex − cos x·ex − I, so 2I = ex(sin x − cos x), and I = (1/2)ex(sin x − cos x) + C. The (1/2) is the small coefficient that catches candidates who forget to divide by 2 after the rearrangement.
Tabulation: a BC-friendly shortcut for repeated applications
When the integrand is a polynomial times a single exponential or trigonometric function, the tabulation method (sometimes called the tabular method, or Diagrammatic Integration by Parts) is faster and less error-prone than repeated application of the formula. AP graders accept it, and BC-level students are explicitly encouraged to learn it because it is harder to misapply on a long chain of applications.
Set up two columns. In the left column, list the derivative of the u-factor, taking derivatives until the column reaches zero. In the right column, list the antiderivative of the dv-factor, taking antiderivatives for as many rows as the left column. Then sum the diagonal products, alternating signs starting with positive, and stop before the row in which the left column is zero. For ∫x2·ex dx, the table looks like this:
| Sign | D (derivatives of u) | I (antiderivatives of dv) |
|---|---|---|
| + | x2 | ex |
| − | 2x | ex |
| + | 2 | ex |
| − | 0 | — |
The answer is the sum of the diagonal products: x2·ex − 2x·ex + 2·ex + C, matching the result from the multi-step application. Tabulation is especially useful when the polynomial is degree three or higher, where four or five applications of the formula are required and the bookkeeping becomes fragile.
Tabulation does not help when the dv factor does not stay in a single family across antiderivatives. The integral ∫sin x·cos x dx is a clean u-substitution, not a tabulation problem. The integral ∫x·ln x dx requires integration by parts but not tabulation, because the Ln x factor does not have a tidy antiderivative family. Knowing the limits of the method is part of the BC-level skill.