AP Calculus BC candidates meet integration by parts as a single formula, ∫u dv = uv − ∫v du, but the College Board rubric scores the technique as a sequence of independent rows on the Free Response section. Every row — the choice of u and dv, the correct computation of du and v, the back-substituted product uv, the new integral of v du, and the final evaluation — can be earned or lost separately. The technique is also tested on the BC-only portion of the exam, which means it appears inside problems that are designed to be more involved than the standard antiderivative prompts. A preparation strategy that treats the formula as one move will leave a row of points on the table; a strategy that treats it as five scored rows, each with its own tactical decision, will pick those points up reliably.
This article walks through the exact structure the rubric uses to award credit for integration by parts on AP Calculus BC FRQs, the four error patterns that most often drop a row, and a preparation plan that targets each scored row in isolation. The goal is to turn a formula students recognise into a procedure they can execute under the six-minute time budget an FRQ subpart typically receives.
Where integration by parts lives on the AP Calculus BC exam and what the rubric is actually measuring
Integration by parts is tested on the AP Calculus BC exam, never on AB. A candidate who has memorised a single statement of the formula has not yet memorised the row structure the reader uses to score the answer, and the row structure is what determines the digit at the end of the score. The College Board publishes a general scoring guideline for integration problems, and for integration by parts the guideline splits the work into at least four independently scored rows: the selection of u and dv from the integrand, the correct computation of du and v, the appearance of the product uv outside the new integral, and the appearance of the new integral of v du in correct form. A fifth row — the final evaluation, including any constant of integration on an indefinite result and the numeric evaluation on a definite result — is scored when the question demands it.
The reason the rubric breaks the work into rows is that a candidate often makes one correct move and one incorrect move on the same subpart. If the rubric awarded the setup point and the final answer point as a single block, a student who chose u correctly but mis-differentiated would lose both rows. Splitting the work means a partial-credit answer is possible, and the preparation plan has to assume that partial credit is the realistic target for the first encounter. In practice, a well-prepared BC candidate earns three to four of those rows on a first attempt and recovers the missing row on a second pass within the same problem. The reader does not care that the candidate knows the formula; the reader cares that each row is independently defensible.
A second reason the row structure matters is that integration by parts is rarely tested in isolation on a BC FRQ. It is embedded inside a longer problem — a differential equation, a volume of revolution, a definite integral that combines with a partner computed by u-substitution, or an improper integral. The exam rewards candidates who can perform the technique in the middle of a longer calculation without losing the surrounding work. The row structure of the rubric, by giving partial credit on each step, is designed to let a candidate who can do everything except the final evaluation still earn a strong score on the subpart. Recognising that the reader is scoring rows, not paragraphs, is the first tactical shift in preparing for this topic.
The u and dv split: the first scored row and how to choose it
The first row of the integration-by-parts rubric is awarded when the candidate writes the integrand as a product of two factors and labels one of them u and the other dv. The reader does not require any particular labelling rule to be applied, but the rule does need to be applied correctly, and the labels must appear on the page. A common failure mode is to write the integrand, then jump straight to the formula with no labels at all. The reader cannot award the setup row because the candidate has not demonstrated the decision the rubric is measuring. Writing u = x and dv = e^x dx on the page costs two seconds and locks in a row of credit.
Two labelling heuristics dominate the published preparation materials. The LIATE rule (logarithmic, inverse trigonometric, algebraic, trigonometric, exponential) tells the candidate to let u be the factor that comes first in the list. The Aden rule, also called the tabular method, ranks the same families but is built for repeated application of the formula. For a one-step application, LIATE is faster and is what most preparation books recommend. For an integral that requires integration by parts to be applied twice, the tabular method is faster because the candidate lists the columns of u, du, v, dv and stops when the bottom row is something they can integrate directly. A BC exam problem that requires two applications will state the requirement; the candidate does not need to invent it.
The failure modes on the first row are concrete and predictable. The first is choosing u from the wrong family — for example, letting u = x^2 in ∫x^2 ln x dx, which forces an algebraic polynomial to be differentiated into 2x and an algebraic polynomial to be integrated. The integral gets harder, not easier. The second is forgetting the dx when writing dv, so the reader sees dv = e^x and cannot tell whether the candidate meant to integrate with respect to x. The third is splitting an integrand that is not a product — for example, treating ∫x e^(x²) dx as a u-substitution candidate when the form is closer to a product of two functions of x. A pre-flight check that the integrand is a product, that the dx is on the dv, and that the chosen u is from a family higher in LIATE than the dv, takes five seconds and prevents the first row from being lost.
For most candidates reading this, the single highest-leverage move on the u and dv row is to write the labels on the page. In my experience the row is lost more often to absent labels than to wrong labels. A reader who can see u and dv, and who can verify that the chosen split is defensible, will award the row even if the choice is non-standard.
Worked example: ∫x cos x dx on the first row
Take ∫x cos x dx. The integrand is a product of x and cos x. By LIATE, algebraic comes before trigonometric, so u = x and dv = cos x dx. Writing those two lines on the page locks in the first row. No calculation has been performed yet; the candidate has only labelled the integrand, and that label is what the rubric scores. From here, the next row of work begins.
The du and v row: differentiation and antidifferentiation that must both be correct
Once u and dv are written, the next scored row is the appearance of du and v on the page. du is the differential of u with respect to x. v is the antiderivative of dv. Both are required. A common error is to write du correctly and then attempt v, fail, and erase the attempt — leaving the reader with du but no v. The reader cannot award the row because the candidate has not shown that the antidifferentiation of dv was performed. A second common error is to differentiate the wrong factor: for ∫x cos x dx with u = x, the du row is du = dx, not du = 1 dx. The reader accepts du = 1 dx and du = dx interchangeably, but a du that is unrelated to the derivative of u is a different story.
The v row is where the most arithmetic slips occur. For dv = cos x dx, v = sin x. For dv = e^x dx, v = e^x. For dv = sec^2 x dx, v = tan x. For dv = 1/x dx, v = ln|x|. Each of these is a single-line evaluation, but each is also the place where a sign error or a missing absolute value is introduced, and that error propagates into the next row. The preparation plan should rehearse the standard v pairs as a separate drill, the way a pianist rehearses scales. A candidate who has to think about whether v of cos x is sin x or −sin x is spending the time budget on a row that should be automatic.
The du and v row is also where the BC-only extensions appear. The exam sometimes gives an integrand that requires the candidate to recognise an inverse trigonometric derivative pattern: dv = 1/(1 + x^2) dx implies v = arctan x, and the rubric expects the candidate to remember this without prompting. A second BC-only extension is hyperbolic functions: dv = cosh x dx implies v = sinh x. These pairs are not on the AB syllabus, so a candidate who has studied only AB material will not recognise them and will lose the row. The preparation plan for BC has to include the inverse trigonometric and hyperbolic antiderivatives as standalone drill items, separate from the integration-by-parts formula.
For most candidates, the du and v row is the row where time is best spent on checking, not on original work. The original work is mechanical; the check is whether the differential of u and the antiderivative of dv are both on the page in correct form before the candidate moves on. A self-check takes three seconds and prevents a downstream row from being scored on top of an error.
The uv minus integral row: the product and the new integral in correct form
The third scored row of integration by parts is the appearance of the formula ∫u dv = uv − ∫v du in concrete form, with the candidate's u, v, du written into the slots. The reader wants to see the product uv on the page and the new integral of v du on the page, with the minus sign between them. A candidate who writes only uv and forgets the new integral, or who writes the new integral in terms of u and du instead of v and du, will not earn this row.
Two specific traps appear at this stage. The first is the back-substitution error: the candidate computes v correctly but then writes uv as u times the original dv, not u times the new v. This is a notational slip, not a conceptual one, and the reader will sometimes interpret it generously. A safer move is to write the full expression uv − ∫v du on a separate line, so the reader can see the product and the new integral as two distinct terms. The second trap is the sign error: uv − ∫v du has a single minus sign, and a candidate who writes uv + ∫v du or uv − ∫u dv will lose the row. The minus sign is a small mark on the page, but the rubric is unforgiving about it.
A third trap is specific to BC problems that combine integration by parts with a partner technique. A common BC problem is to compute the integral of a product by splitting it into two integrals, applying integration by parts to one and u-substitution to the other, and then combining the results. In this construction, the integration-by-parts subpart must end with a clean expression for the new integral of v du, because the next subpart will use that expression as a building block. A candidate who leaves the new integral in unevaluated form on the page is doing exactly what the problem requires; a candidate who tries to finish the evaluation inside the integration-by-parts subpart is doing more work than the rubric rewards and may introduce an error in the building block.
For most candidates reading this, the highest-leverage move on the uv minus integral row is to keep the product and the new integral on separate visual lines. A reader scanning the page can confirm the row in two seconds if the layout is clear, and a candidate can confirm their own work in two seconds if the layout is clear. The formula is one line, but the row the rubric scores is two: the product, and the new integral.
The final evaluation row: definite integrals, the constant of integration, and the partner integral
The final row of integration by parts is the evaluation of the result. The form of the evaluation depends on whether the original integral was definite or indefinite. For an indefinite integral, the result is uv − ∫v du plus a constant of integration. For a definite integral, the result is a numeric value obtained by evaluating the new expression at the bounds. The rubric scores the constant of integration on the indefinite row and the numeric evaluation on the definite row, and a candidate who omits either will lose the corresponding row.
The constant of integration is a frequent point of loss on indefinite integration by parts. The exam question will sometimes direct the candidate to find a particular antiderivative satisfying an initial condition, in which case the constant is determined by the condition and the +C is not required. More often, the question asks for the most general antiderivative, and the +C is the marker that the candidate understood the form of the answer. A candidate who writes a clean antiderivative without the +C will lose the row, even though the antiderivative is correct. The +C is a single character on the page, and it is the difference between a five and a four on the subpart.
On a definite integral, the evaluation has two sub-rows. The first is the substitution of the bounds into the new expression uv − ∫v du. The second is the arithmetic of the evaluation. A candidate who substitutes the bounds correctly but makes an arithmetic slip on the subtraction will lose the second sub-row but may retain the first. The reader is looking for a final numeric answer, and a numeric answer that is off by a sign because of an arithmetic slip is a partial-credit situation, not a zero.
The partner-integral construction, common on BC FRQs, adds a third sub-row. When the original problem is split into two parts, one computed by integration by parts and the other by u-substitution, the final row of the integration-by-parts subpart is the expression for the new integral of v du, written so the next subpart can use it. The reader scores this row as the integration-by-parts row, but the value of the row is that it sets up the next subpart. A preparation plan that rehearses the partner-integral construction as a single problem, not as two separate problems, will catch the connection between the subparts and prevent the candidate from leaving the building block in an unevaluated form when the problem expects it evaluated.
Common pitfalls and how to avoid them. First, the absent-label pitfall: write u and dv on the page even if the labels feel obvious. Second, the missing-absolute-value pitfall: when v is ln|x|, write ln|x|, not ln x. Third, the sign-slip pitfall: the formula has a single minus sign; double-check it before moving on. Fourth, the partner-integral pitfall: when the problem is split, leave the new integral in clean form for the next subpart. Fifth, the constant-of-integration pitfall: write +C on an indefinite result unless an initial condition is given. Each pitfall costs a row, and the rows are independent. A preparation plan that targets each pitfall as a separate drill item will cut the row losses on the exam.
Four integration-by-parts failure modes the AP Calculus BC rubric catches
Across published FRQs and rubric commentary, four failure modes appear often enough to warrant their own preparation drills. The first is the circular-integral failure: the candidate chooses u and dv in a way that produces a new integral that is identical to the original. This happens most often with integrands of the form e^x sin x or e^x cos x, where the choice of u matters. The remedy is to let u be the trigonometric factor and dv be the exponential factor, so the second application of the formula brings the trigonometric factor back to its original form, allowing the two integrals to be combined. A candidate who picks the wrong direction has to restart the problem.
The second is the back-substitution failure: the candidate computes v correctly but then writes the new integral in terms of u and du instead of v and du. The reader sees a row that is not in the form the rubric is scoring, and the row is lost. The remedy is to write the new integral on a separate line, in the slot of the formula, so the substitution is mechanical and the form is obvious.
The third is the missing-differential failure: the candidate writes du = 2x without the dx, or v = sin x without the implicit reference to the original dv. The reader cannot tell whether the candidate understood that du is a differential and v is an antiderivative. The remedy is to write du = 2x dx and v = sin x, even though the dx is implied, because the explicit form is what the rubric scores.
The fourth is the premature-evaluation failure: the candidate attempts to evaluate the new integral of v du inside the integration-by-parts subpart, before the partner integral or the next step of the problem is set up. This is most costly on the partner-integral construction, where the new integral is a building block for the next subpart. The remedy is to stop at the formula line, write the new integral cleanly, and wait for the next subpart to direct the evaluation. The reader will score the formula line and the clean new integral as two separate rows, and the next subpart will use that clean form.
AP Calculus BC FRQ question types that embed integration by parts
Three question types appear often enough on the BC FRQ section to be treated as the canonical containers for integration by parts. The first is the differential-equation question, where the candidate is given a first-order linear differential equation and asked to find a particular solution satisfying an initial condition. The standard form requires multiplying by an integrating factor, which is computed by integrating a coefficient — and the coefficient is often chosen so that the integrating factor is the product of two functions whose antiderivative requires integration by parts. The integration-by-parts subpart is typically the second or third part of a multi-part question, and the result feeds into the next subpart.
The second is the volume-of-revolution question, where the candidate is asked to compute a volume by the disk or shell method, and the resulting integral requires integration by parts. BC problems will sometimes pair the disk method on one part of the figure with the shell method on another part, and the integration-by-parts subpart will appear inside one of the two method calculations. The rubric for the subpart is the standard four-row structure, and the result of the subpart is the volume itself.
The third is the improper-integral question, where the candidate is asked to evaluate an integral with an infinite bound or a discontinuity, and the integrand is a product that requires integration by parts. The BC exam tests improper integrals more often than AB, and the integration-by-parts subpart is the inner step that the limit calculation depends on. The rubric for the subpart is again the standard four-row structure, and the result is the limit value.
For most candidates reading this, the most effective preparation move is to drill each of these three question types as a separate case. The integration-by-parts technique is the same in all three, but the surrounding work — the differential equation, the volume, the improper-integral limit — is different, and the candidate who has rehearsed each case will spend less time orienting and more time on the scored rows.
Comparison: integration by parts versus u-substitution on the BC FRQ
The two techniques are sometimes confused, especially on partner-integral problems where both appear. The table below summarises the decision points the rubric uses to score each.
| Feature | Integration by parts | u-substitution |
|---|---|---|
| Integrand form | Product of two functions of x from different families | Composite function: f(g(x)) g'(x) |
| First scored row | Labels u and dv on the page | Choice of u, with du computed |
| Back-substitution | New integral in terms of v and du | Integral in terms of u and du, then bounds in u |
| Common trap | Circular integral when families are mis-chosen | Forgetting to convert bounds, leaving limits in x |
| Rubric row count | Four to five rows | Three to four rows |
| BC-only elements | Inverse trig and hyperbolic v pairs | Long-division setup before substitution |
Integration by parts on the AP Calculus BC exam: a six-row preparation plan
A preparation plan that targets the rows of the rubric, rather than the formula in the abstract, will produce the strongest score gain for the time invested. The plan has six components, each tied to a row or a preparation stage.
First, drill the LIATE families as a standalone exercise. The candidate should be able to look at an integrand of the form x^n e^x, x^n sin x, x^n cos x, x^n ln x, e^x sin x, e^x cos x, and identify which factor is u and which is dv in under ten seconds. This drill is mechanical, takes a single sitting, and pays off across the entire topic.
Second, drill the v pairs. The candidate should be able to state v for dv = e^x, sin x, cos x, sec^2 x, 1/x, 1/√(1−x^2), and 1/(1+x^2) without hesitation. The BC-only pairs — 1/√(x^2−1) and the hyperbolic functions — should be added to the drill if the candidate is preparing for BC.
Third, drill the row layout. The candidate should write the four rows of the rubric — u and dv, du and v, uv and the new integral, and the final evaluation — on a single practice problem and verify that each row is visible on the page. The layout is the difference between a reader awarding credit and a reader searching for the row.
Fourth, drill the partner-integral construction. The candidate should work a problem that splits an integral into two parts, one computed by integration by parts and the other by u-substitution, and verify that the integration-by-parts subpart ends with a clean expression for the new integral that the u-substitution subpart can use. This is the construction that most often loses the building-block row.
Fifth, drill the definite-integral evaluation. The candidate should work a definite integration-by-parts problem, evaluate uv − ∫v du at the bounds, and verify the arithmetic. Sign slips are the most common error here, and a self-check that involves re-deriving the sign on a separate line prevents the loss.
Sixth, drill the +C on indefinite results. The candidate should work an indefinite integration-by-parts problem and verify that the final line includes +C unless an initial condition is given. The +C is a single character, and the drill is to write it on every indefinite result as a default.
For most candidates, the six drills together take three to four hours of focused practice, and they cover the row structure of the rubric end to end. After the drills, the candidate can approach any integration-by-parts problem on the BC FRQ section with a procedure that targets the rows rather than relying on the formula in the abstract.
Reading the rubric, not the formula: a final tactical note on AP Calculus BC scoring
The single largest shift a candidate can make in preparing for integration by parts on the AP Calculus BC exam is to stop studying the formula and start studying the rubric. The formula is one line. The rubric is four to five rows. The exam scores the rows, not the formula, and a candidate who has internalised the row structure will earn partial credit on a problem where the technique breaks down, will lose fewer rows to absent labels, and will recover from a sign slip without losing the setup. The preparation plan above targets the rows; the drills above rehearse the rows; the exam rewards the rows. For a candidate who treats integration by parts as a scored sequence rather than a single move, the topic moves from a perceived weak spot to a reliable source of points.
For candidates who want a structured pass through the row-by-row preparation plan, with worked FRQ problems and a six-hour drill schedule, AP Courses' one-to-one AP Calculus BC programme pairs each candidate with a tutor who scores their integration-by-parts attempts against the actual rubric rows, identifies the row the candidate most often loses, and builds a targeted drill sequence around that row. The result is a preparation plan that turns the integration-by-parts formula into a scored sequence on the BC FRQ section.