Integration by Parts

Last Updated: July 2026

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Integration by parts is a technique for integrating the product of two functions — the reverse of the product rule for differentiation.

This A-Level guide covers the formula and its proof, how to choose u and dv/dx, repeated and cyclic applications, and five exam-style questions with worked solutions. Download the free PDF worksheet or watch the video tutorial to teach or revise the topic.

Published: June 2026  |  Last updated: July 2026

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What is integration by parts?

Integration by parts lets you integrate a product of two functions by relating it to a simpler integral. It is the integration counterpart of the product rule: If y = uv then

\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}

Rearranging and integrating both sides gives the integration by parts formula.

At A-Level, integration by parts is used for products such as a polynomial multiplied by an exponential or trigonometric function, logarithmic integrals like ∫ ln x dx, and more advanced types including cyclic integrals where the original integral reappears after two applications.

The integration by parts formula

The standard formula is:

\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx

Teacher’s Guide

It is important for students to derive the integration by parts formula. Too often, the formula is given with little explanation of where it comes from and how it is a natural extension of the product rule of differentiation. In my experience, students really appreciate taking the time to prove the formula as it helps them understand its usefulness when finding the integral of two products of x.

As we work through the examples in the lesson pack we take the time to discuss which function of x should map onto the u or dv/dx term. Taking the time at the start of the question saves a lot of time and effort later on.

When students can apply the formula for basic integrals we quickly move on to integrating functions that need to be integrated more than once, for example:

∫x^2sinx dx

and using integration by parts to find the area under a curve. There are some really interesting integrals that can be evaluated through integration by parts — my favourites are included in the exam style questions below.

I give students a printed handout from the lesson pack of the example I am modelling. Working alongside rather than copying from the board keeps the focus on choosing u and dv/dx correctly before any algebra begins. By the end of a worked example, students have a correctly annotated solution to reference during independent practice.

Video: integration by parts

Watch the tutorial below to see integration by parts modelled step by step: deriving the formula from the product rule, choosing u and dv/dx, and working through examples including repeated and cyclic applications.

Want to practise alongside the video? Download the lesson pack and pause after each example to try the next question before the worked solution.

What the video covers

• Deriving the formula — starting from the product rule d(uv)/dx = u(dv/dx) + v(du/dx) and rearranging to obtain ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx.
• Choosing u and dv/dx — using the LATE or LIATE acronym (Logarithmic, Algebraic, Trigonometric, Exponential — with Inverse trig added for LIATE) to decide which part of the product becomes u.
• Basic application — integrating a polynomial multiplied by an exponential, such as ∫ (5x − 3)e^3x dx.
• Repeated and cyclic integration by parts — applying the formula twice for integrals like ∫ (e^x sin x) dx, where the original integral reappears and can be collected algebraically.

The written step-by-step guide below follows the same order as the video. Use the video for live modelling in class, or the text sections for independent revision.

How to use integration by parts (step by step)

Follow these four steps to integrate any product using integration by parts. The worked example integrates ∫ (5x − 3)e^3x dx, matching Question 1 in the exam section below.

Step 1: Choose u and dv/dx

Identify the two factors in the integrand. You can use the LATE or LIATE acronym to easily remember the order of priority for choosing u: Logarithmic, Algebraic, Trigonometric, Exponential (LIATE also includes Inverse trig before Algebraic). Choose which factor becomes u (the part you differentiate) and which becomes dv/dx (the part you integrate). The goal is to make the remaining integral ∫ v (du/dx) dx simpler than the original.

For ∫ (5x − 3)e^3x dx: let u = 5x − 3 (algebraic) and dv/dx = e^3x (exponential).

Step 2: Differentiate u and Integrate dv/dx

Find du/dx by differentiating u, and find v by integrating dv/dx. Do not add a constant of integration when finding v for an indefinite integral — the constant is absorbed into C at the end.

Step 3: Apply the formula

Substitute into ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx:

∫ (5x − 3)e^3x dx = (5x − 3)(⅓e^3x) − ∫ ⅓e^3x · 5 dx

Step 4: Evaluate the new integral

The remaining integral should be straightforward. Integrate, simplify, and add C for indefinite integrals.

= ⅓(5x − 3)e^3x − ⁵⁄₉e^3x + C = ^{(15x − 14)e3x}⁄₉ + C

How to apply integration by parts

Once the basic four-step method is secure, A-Level questions test three harder variations: applying the formula more than once, cyclic integrals where the original reappears, and definite integrals used to find areas.

Repeated integration by parts

When the new integral ∫ v (du/dx) dx is still a product, apply integration by parts again. Each application should simplify the algebraic part of u. For ∫ x² sin x dx, two applications reduce x² to a constant.

For ∫ (ln x)² dx, the first application gives a term involving ∫ ln x dx, which itself requires a second application of integration by parts (with u = ln x, dv/dx = 1).

Cyclic integrals

Some integrals, such as ∫ e^x sin x dx, produce the original integral after two applications of integration by parts. Label the first result as equation (1), apply parts again to the new integral to get equation (2), then substitute (2) into (1). The original integral appears on both sides — collect terms and solve algebraically.

Definite integrals and area

For definite integrals, evaluate [ uv ] between the limits before integrating the remaining term. When finding a shaded area bounded by a curve and a normal line, split the region into parts: the area under the curve (found by definite integration) plus any triangular or trapezoidal section found using coordinate geometry.

Common mistakes with integration by parts

• Choosing u and dv/dx the wrong way round — if the new integral is harder than the original, swap your choices and start again.
• Forgetting the minus sign in the formula — it is uv minus ∫ v (du/dx) dx, not plus.
• Adding a constant when finding v — only add C at the very end of an indefinite integral.
• Sign errors when integrating trigonometric functions — ∫ sin x dx = −cos x, not +cos x. A sign error here propagates through the entire solution.
• Not collecting cyclic terms — in integrals like ∫ e^x sin x dx, students apply parts twice but forget to bring the repeated integral to one side and solve.
• Dropping the boundary term in definite integrals — remember to evaluate uv at both limits before integrating the remaining part.
• Stopping after one application when two are needed — integrals involving (ln x)² or x² sin x require a second application of the formula.

A-Level integration by parts checklist

Use this list to check you can tackle any integration by parts question on an A-Level paper.

• I can state the integration by parts formula from memory.
• I can derive the formula from the product rule for differentiation.
• I can use the LATE or LIATE acronym to choose u and dv/dx correctly.
• I can integrate a polynomial multiplied by an exponential or trigonometric function.
• I can integrate logarithmic functions such as ∫ ln x dx and ∫ (ln x)² dx.
• I can apply integration by parts more than once when required.
• I can solve cyclic integrals by collecting the repeated integral algebraically.
• I can evaluate definite integrals using integration by parts, including the boundary term.
• I can use integration by parts to find the area of a region bounded by a curve and a straight line.
• I can express an answer in a required form, such as ^{(px − q)erx}⁄_s + C.

Free integration by parts worksheet (PDF)

Work through these five A-Level exam-style questions. Select each accordion to reveal the worked solution. The same questions are in the free PDF worksheet.

Exam Style Questions

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Frequently asked questions

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