A-Level Mathematics: Mastering Integration by Parts

Integration by parts is one of those calculus techniques that students either master or avoid forever. The formula looks weird, it doesn't feel intuitive, and when you get it wrong (which you will, repeatedly), you end up with integrals that are harder than the one you started with. But here's the secret: integration by parts isn't mysterious. It's just the product rule in reverse, and once you see it that way, it clicks. (This guide has been for the 2025-26 syllabus.)

This guide walks through the intuition behind integration by parts, shows you how to recognize when to use it, and gives you the systematic approach that separates students who fumble through it from students who nail every problem.

Whether you're preparing for A-Level exams or working toward scoring a 7 in IB Math HL, integration by parts is a technique you need to master.

What Integration by Parts Actually Is

Before learning the formula, understand the concept. Differentiation has a product rule: if you have two functions multiplied together, you differentiate one and leave the other, then add the derivative of the second times the first. Integration by parts is the inverse of that. For more on this, see our guide on understanding IB grading.

The product rule for derivatives: d/dx[u(x)·v(x)] = u'(x)·v(x) + u(x)·v'(x)

Rearranged: u(x)·v'(x) = d/dx[u(x)·v(x)] - u'(x)·v(x)

Integrated: ∫u(x)·v'(x)dx = u(x)·v(x) - ∫u'(x)·v(x)dx

This is integration by parts. You're trading one integral for a different integral plus a product term. The hope is that the new integral is easier to solve than the original.

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Why It Works: The Intuition

The key insight is that integration by parts lets you "move" the differentiation from one function to another. If you have a product like x·sin(x), differentiating x gives you 1 (simpler), but differentiating sin(x) gives you cos(x) (not simpler). By using integration by parts, you differentiate x and integrate sin(x) instead, ending up with a simpler integral.

That's the goal: transform a hard integral into an easier one by choosing which function to differentiate and which to integrate.

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The LIATE Rule: How to Choose Your u and dv

The integration by parts formula requires you to split your integrand into two parts: u and dv. Choosing correctly is half the battle. Use LIATE:

L: Logarithmic functions (ln, log)
I: Inverse trigonometric functions (arcsin, arccos, arctan)
A: Algebraic functions (polynomials like x, x², etc.)
T: Trigonometric functions (sin, cos, tan)
E: Exponential functions (e^x, a^x)

The rule: Your u should be whichever function appears first in LIATE. The rest (including dx) is dv.

Example: ∫x·sin(x)dx

• You have an algebraic function (x) and a trig function (sin(x)).
• Algebraic comes before Trigonometric in LIATE, so u = x and dv = sin(x)dx.
• Then: du = 1·dx and v = -cos(x).
• Applying the formula: ∫x·sin(x)dx = x·(-cos(x)) - ∫(-cos(x))·1·dx = -x·cos(x) + ∫cos(x)dx = -x·cos(x) + sin(x) + C.

The algebraic function was easier to differentiate (it became 1), and we integrated the trig function (which we can do easily). Result: a solvable integral.

When LIATE Doesn't Work

LIATE is a heuristic, not a law. Sometimes you'll try it and end up with a harder integral. If that happens, try the opposite assignment (swap u and dv) or recognize that this integral might need a different technique altogether.

The Integration by Parts Process, Step by Step

Step 1: Identify that you need integration by parts. Your integrand is a product of two different types of functions (like x·e^x or ln(x)·x²). It's not something you can do with u-substitution or basic integration rules.

Step 2: Assign u and dv using LIATE. Identify which function comes first in LIATE. That's u. Everything else is dv.

Step 3: Find du and v. Differentiate u to get du. Integrate dv to get v.

Step 4: Plug into the formula. ∫u·dv = u·v - ∫v·du

Step 5: Evaluate ∫v·du. This should be simpler than the original integral. If it's not, you may have assigned u and dv wrong.

Step 6: Add the constant of integration. Don't forget +C if this is an indefinite integral.

Common Integration by Parts Problems and How to Solve Them

Type 1: Polynomial × Exponential (like x·e^x)

Strategy: Use u = polynomial, dv = e^x·dx.

Why: Differentiating the polynomial reduces its degree. Integrating e^x is easy and stays e^x.

Example: ∫x²·e^x·dx

• u = x², du = 2x·dx
• dv = e^x·dx, v = e^x
• ∫x²·e^x·dx = x²·e^x - ∫e^x·2x·dx = x²·e^x - 2∫x·e^x·dx
• Now you have ∫x·e^x·dx, which you need to solve by integration by parts again (called repeated integration by parts). This technique of repeated integration by parts also appears frequently in AP Calculus BC series and sequences problems.

Type 2: Polynomial × Trig (like x·sin(x))

Strategy: Use u = polynomial, dv = trig·dx.

Why: Same as above — differentiate the polynomial, integrate the trig function.

Example: ∫x·cos(x)·dx

• u = x, du = 1·dx
• dv = cos(x)·dx, v = sin(x)
• ∫x·cos(x)·dx = x·sin(x) - ∫sin(x)·1·dx = x·sin(x) + cos(x) + C

Type 3: Logarithmic × Polynomial (like ln(x)·x)

Strategy: Use u = ln(x), dv = polynomial·dx.

Why: Logarithmic functions differentiate to 1/x (much simpler). Integrating polynomials is trivial.

Example: ∫ln(x)·x·dx

• u = ln(x), du = 1/x·dx
• dv = x·dx, v = x²/2
• ∫ln(x)·x·dx = ln(x)·(x²/2) - ∫(x²/2)·(1/x)·dx = (x²·ln(x))/2 - ∫(x/2)·dx = (x²·ln(x))/2 - x²/4 + C

Type 4: Trig × Exponential (like sin(x)·e^x)

Strategy: Use u = trig, dv = e^x·dx (or vice versa). This one is special because you may need to apply integration by parts twice and solve algebraically.

Example: ∫e^x·sin(x)·dx

• u = sin(x), du = cos(x)·dx
• dv = e^x·dx, v = e^x
• ∫e^x·sin(x)·dx = e^x·sin(x) - ∫e^x·cos(x)·dx
• Now integrate by parts again on ∫e^x·cos(x)·dx:
• u = cos(x), du = -sin(x)·dx
• dv = e^x·dx, v = e^x
• ∫e^x·cos(x)·dx = e^x·cos(x) - ∫e^x·(-sin(x))·dx = e^x·cos(x) + ∫e^x·sin(x)·dx
• Now substitute back: ∫e^x·sin(x)·dx = e^x·sin(x) - [e^x·cos(x) + ∫e^x·sin(x)·dx]
• Solve for ∫e^x·sin(x)·dx algebraically to get: ∫e^x·sin(x)·dx = (e^x·(sin(x) - cos(x)))/2 + C

Common Mistakes and How to Avoid Them

1. Choosing u and dv incorrectly. If your new integral is harder than the original, you chose wrong. Use LIATE and trust the system. If it doesn't work, the integral probably needs a different technique. Building strong problem-solving instincts takes practice—students preparing for AP Calculus AB face similar challenges with technique selection, and understanding systematic learning approaches across subjects strengthens your mathematical reasoning.

2. Forgetting the negative sign in the formula. The formula is ∫u·dv = u·v - ∫v·du. That minus sign is crucial. Forgetting it flips your answer.

3. Not recognizing when to use integration by parts. You need a product of different function types. If you have u·u' (like x·2x or sin(x)·cos(x)), use u-substitution instead, not integration by parts.

4. Making algebra errors in the setup. After choosing u and dv, find du and v carefully. A mistake here propagates through the entire problem.

5. Forgetting +C on indefinite integrals. Every indefinite integral needs +C. It's not optional.

6. Stopping too early. After applying the formula, you may get ∫v·du, which still needs to be integrated. Don't leave it there — finish the job.

When to Use Integration by Parts vs. Other Techniques

Integration by parts: Product of two different function types (polynomial × trig, polynomial × exponential, logarithm × polynomial, etc.).

U-substitution: Chain rule in reverse. You have a composition of functions (like sin(3x) or e^(x²)), not a product.

Partial fractions: Rational functions (polynomials divided by polynomials).

Trigonometric identities: Integrals involving only trig functions, especially powers of sin and cos.

Pattern recognition: Some integrals (like ∫1/(x²+1)·dx) match standard forms you should memorize.

If you're unsure which technique to use, developing a systematic approach like the one outlined in our IB Math AA HL calculus guide can help. Ask: Is this a product? If yes, try integration by parts. Is it a composition (something inside something else)? If yes, try u-substitution. Is it a fraction of polynomials? If yes, try partial fractions. Understanding systematic study approaches for mastering mathematical techniques applies across all technical subjects.

Master Integration by Parts with Targeted Practice

Work with an A-Level maths tutor who can diagnose where your understanding breaks down and coach you through the trickiest integration by parts problems → Integration by parts is learnable, but it takes practice and feedback. Our tutors teach you the conceptual foundation (so it's not just a formula you memorize), help you recognize when to use it, and give you targeted feedback on your solution method. Whether you're building confidence on the basics or refining your technique for exam problems, expert support accelerates your progress.

FAQs

Why is there a minus sign in the integration by parts formula?

The minus sign comes from rearranging the product rule. When you differentiate u·v, you get u'·v + u·v'. Rearranging to isolate u·v' gives you u·v' = d(u·v) - u'·v. Integrating both sides introduces the minus sign.

What if I apply integration by parts and end up with an even harder integral?

You likely assigned u and dv incorrectly. Try reversing your choice (make the other function your u). If it's still harder, this integral probably needs a different technique, not integration by parts.

How do I know if I need integration by parts or u-substitution?

Integration by parts is for products of different types of functions. U-substitution is for compositions (something nested inside something else). If you see f(x)·g(x) (product), lean toward integration by parts. If you see f(g(x)) (composition), lean toward u-substitution.