Integration by parts feels like guesswork because the formula does not tell you how to divide the integrand into u and dv. That choice is strategic: differentiate the factor that becomes simpler, and integrate the factor whose antiderivative remains manageable. Once you judge a choice by the new integral it creates, the method becomes a controlled transformation rather than trial and error.
For current IB Mathematics: Analysis and Approaches, integration by parts appears in AHL 5.16, so it is required content for Maths AA HL, not AA SL. The official guide includes integration by parts, repeated integration by parts, and examples such as ∫x sin x dx, ∫ln x dx, ∫x²eˣ dx, and ∫eˣ sin x dx. This article focuses specifically on choosing u and dv; for the wider syllabus context, use RevisionDojo’s IB Maths AA Calculus Explained resources.
What the integration by parts rule actually does
The integration by parts rule is
∫u dv = uv - ∫v du.
Equivalently,
∫u(x)v′(x) dx = u(x)v(x) - ∫v(x)u′(x) dx.
It comes directly from the product rule:
d(uv)/dx = u(dv/dx) + v(du/dx).
Rearranging and integrating gives the integration by parts formula. It is therefore best understood as the product rule used backwards, but with an important qualification: it does not directly produce the answer. It replaces one integral with another integral, and your choice is successful only when the replacement is easier.
This is the central logic:
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u is differentiated to obtain
du. -
dv is integrated to obtain
v. -
The new integral is
∫v du. -
Your aim is to make
∫v dusimpler than the original.
Students often focus on remembering the letters while overlooking this purpose. The letters are merely bookkeeping; the simplification of the new integral is the mathematical objective.
Why choosing u and dv feels uncertain
Differentiation rules usually tell you exactly what to do. If you see a product, you apply the product rule; if you see a composite function, you consider the chain rule. Integration is less mechanical because several techniques may be possible, and a formally valid method may still make the expression worse.
For example, consider
∫x eˣ dx.
There are two apparent choices:
ChoiceudvWhat happens?Productivexeˣ dxDifferentiating x gives 1, so the new integral is simpler.Unproductiveeˣx dxIntegrating x gives x²/2, so the new integral becomes more complicated.
Both divisions are legally possible. Only the first moves the problem towards an elementary integral.
With u = x and dv = eˣ dx, we obtain
du = dx and v = eˣ.
Therefore,
∫x eˣ dx = xeˣ - ∫eˣ dx = xeˣ - eˣ + C.
The choice was not based on intuition alone. It was based on predicting that differentiating x would remove the algebraic factor while integrating eˣ would leave its complexity unchanged.
The two-question test for choosing u and dv
Before applying the rule, ask two questions.
1. Which factor improves when differentiated?
A strong choice for u often becomes shorter, lower in degree, or otherwise easier after differentiation:
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x³ → 3x² → 6x → 6 → 0 -
ln x → 1/x -
arctan x → 1/(1+x²) -
arcsin x → 1/√(1-x²)
A polynomial is especially useful as u because repeated differentiation eventually produces zero. Logarithmic and inverse trigonometric functions are useful for a different reason: their derivatives are usually more familiar to integrate than the original functions themselves.
2. Can the remaining factor be integrated cleanly?
The proposed dv must have an antiderivative you can find. Common productive choices include:
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eˣ dx -
sin x dxorcos x dx -
xⁿ dx -
a constant multiple of one of these
Do not choose dv = ln x dx merely because it is the remaining factor. Finding the antiderivative of ln x already requires integration by parts, so the choice has not simplified anything.
A compact decision rule is:
Choose u so differentiation improves it, and choose dv so integration does not damage it.
You should then mentally preview v du. If that product is easier than the original integrand, proceed.
How LIATE helps without becoming another guessing rule
LIATE is a common heuristic for selecting u:
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Logarithmic functions
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Inverse trigonometric functions
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Algebraic functions
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Trigonometric functions
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Exponential functions
Choose as u the factor that appears earliest in the list. The remaining factor, together with dx, becomes dv.
LIATE works often because logarithmic and inverse trigonometric functions are usually easier to differentiate than to integrate, while trigonometric and exponential functions are usually easy to integrate repeatedly. Algebraic functions sit between these groups because differentiation lowers polynomial degree.
IntegralLIATE choice for uChoice for dvReason∫x cos x dxxcos x dxThe polynomial disappears after differentiation.∫x²eˣ dxx²eˣ dxRepeated differentiation reduces the polynomial to zero.∫x ln x dxln xx dxThe logarithm is easier to differentiate than integrate.∫eˣ sin x dxEither factor can workThe other factorNeither becomes simpler, so a cyclic method is needed.
LIATE is not an official IB rule or a theorem. It is a useful memory aid, and some textbooks use the variation ILATE. If LIATE conflicts with the two-question test, trust the actual algebra: check whether dv can be integrated and whether ∫v du is an improvement.
Hidden products: when the second factor is 1
Some integration by parts questions do not initially look like products. The standard example is
∫ln x dx.
Rewrite it as
∫(ln x)(1) dx.
Choose u = ln x and dv = dx. Then
du = (1/x) dx and v = x.
Applying the rule gives
∫ln x dx = x ln x - ∫x(1/x) dx
= x ln x - x + C.
The same hidden-product idea applies to integrals such as ∫arctan x dx and ∫arcsin x dx. This is why simply asking “Is the integrand visibly a product?” is too restrictive. A better recognition question is: Would differentiating this difficult function make it easier if I pair it with 1?
Repeated integration by parts
Consider
I = ∫x²eˣ dx.
Choose u = x² and dv = eˣ dx, giving du = 2x dx and v = eˣ. Then
I = x²eˣ - 2∫x eˣ dx.
The remaining integral is easier, but it still contains a product. Apply integration by parts again with u = x and dv = eˣ dx:
∫x eˣ dx = xeˣ - eˣ.
Therefore,
I = x²eˣ - 2(xeˣ - eˣ) + C
= eˣ(x² - 2x + 2) + C.
Notice the consistent strategy. The polynomial remains u each time, so its degree falls from 2 to 1 to 0. Switching the roles during the second application would reverse that progress.
For polynomial products with exponential or trigonometric functions, a tabular layout can shorten repeated working. However, the table is not a different theorem; it is compressed repeated integration by parts. You should first understand why differentiating the polynomial is productive before relying on the shortcut.
RevisionDojo’s AHL 5.16 integration notes and AHL 5.16 video lessons provide further worked examples of repeated applications.
Cyclic integration by parts
Sometimes neither factor becomes simpler. A key IB Maths AA HL example is
I = ∫eˣ sin x dx.
If u = sin x and dv = eˣ dx, then
I = eˣ sin x - ∫eˣ cos x dx.
Apply integration by parts again to the remaining integral, maintaining the same general assignment: differentiate the trigonometric factor and integrate the exponential factor.
∫eˣ cos x dx = eˣ cos x + ∫eˣ sin x dx.
The original integral has returned. Substitute it as I:
I = eˣ sin x - eˣ cos x - I.
Hence,
2I = eˣ(sin x - cos x)
and
I = [eˣ(sin x - cos x)]/2 + C.
Here, success does not mean making each intermediate integral visibly simpler. Success means creating a cycle that returns to the original integral, after which algebra isolates it. Naming the original integral I at the start makes this structure much easier to recognize and reduces transcription errors.
Integration by parts or substitution?
Not every product requires integration by parts. In IB Maths AA calculus, substitution should be checked first when one factor is related to the derivative of an inner function.
Integral structureLikely methodRecognition clue∫x(x²+1)⁵ dxSubstitutionx is proportional to the derivative of x²+1.∫x e^(x²) dxSubstitutionThe exponent has derivative 2x.∫x eˣ dxIntegration by partsNo inner function has the other factor as its derivative.∫ln x dxIntegration by partsA difficult single function can be paired with 1.∫x² cos x dxRepeated integration by partsDifferentiation steadily lowers the polynomial degree.
Before using parts, ask whether the integrand contains a clear reverse-chain-rule pattern. Integration by parts is usually appropriate when differentiation of one factor and integration of the other create a simpler product, not merely whenever two expressions are multiplied.
Definite integrals and exam presentation
For a definite integral, the rule becomes
∫[a to b] u dv = [uv] from a to b - ∫[a to b] v du.
Keep the original limits because the variable remains x; unlike substitution, integration by parts does not normally require transformed limits. Apply the boundary values to the complete uv term and to an antiderivative of the remaining integral. Do not include + C in a definite integral.
The current IB guide requires students to use a clean mathematics formula booklet in examinations, but access to a formula does not remove the need to show a valid method. The official specimen materials state that correct answers are not necessarily awarded full marks without supporting working, while correct method can earn partial credit even if the final answer is wrong.
A clear exam solution should show:
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The assignments
u = ...anddv = .... -
The resulting
du = ... dxandv = .... -
A correct substitution into
∫u dv = uv - ∫v du. -
Any repeated or cyclic application.
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+ Cfor an indefinite integral, or correctly applied bounds for a definite integral. -
A simplified final answer where appropriate.
Common mistakes that make the method seem harder
Choosing dv without checking that it can be integrated
Students sometimes select an excellent-looking u but leave an impossible or unhelpful expression as dv. Both choices must work together. Always calculate or at least anticipate v before committing.
Forgetting that dv includes dx
Write dv = eˣ dx, not merely dv = eˣ. This keeps the differential notation consistent and makes the transition to v = eˣ clear.
Losing the minus sign
The formula contains uv - ∫v du. If the new integral itself produces a negative antiderivative, brackets are useful because two negative signs may combine.
Adding a constant too early
You do not need a separate constant each time you integrate dv. Write one + C at the end of the complete indefinite integral. Constants introduced during intermediate applications are absorbed into the final arbitrary constant.
Using LIATE without previewing the result
LIATE suggests a choice; it does not prove that the choice is useful. Mentally form v du and check whether the new integral is simpler, terminates after repetition, or returns cyclically.
Stopping when the original integral returns
In a cyclic problem, the return of the original integral is the intended breakthrough. Name it I, move all terms involving I to one side, and solve algebraically.
How to make the choice automatic in revision
Practise recognition separately from calculation. Take a set of integrals and, before evaluating any of them, record only the likely method, the proposed u, the proposed dv, and what will happen to the new integral. This trains strategic selection rather than memorized execution.
A useful three-stage routine is:
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Stage 1: Classify the structure as inspection, substitution, integration by parts, or another known method.
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Stage 2: Predict whether the new integral will simplify, terminate, or cycle.
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Stage 3: Complete the algebra and verify by differentiating the answer.
Use the AHL 5.16 Questionbank for focused practice, then move to the broader IB Maths AA calculus Questionbank so you must choose between techniques without being told the topic. The AHL 5.16 Flashcards can reinforce the rule and recognition patterns, while Jojo AI can help identify why a particular choice of u created a worse integral.
Once the method is reliable in isolation, test it under mixed conditions using IB Maths AA Predicted Papers. This matters because an examination may not tell you which integration technique to use.
Conclusion
Integration by parts stops feeling like guesswork when you treat it as a decision about the next integral. Choose u because differentiation improves it, choose dv because integration keeps it manageable, and use LIATE only as a supporting heuristic. Repeated problems should show steady simplification, while cyclic problems should deliberately recreate the original integral.
For revision, begin with the RevisionDojo AHL 5.16 notes and Flashcards, practise individual choices in the Questionbank, and then use Jojo AI or timed Predicted Papers to test whether you can recognize the method independently.