Integration by parts is the reverse of the product rule for differentiation, used when a function you need to integrate is a product of two simpler functions that cannot be integrated directly. It appears throughout A-Level Pure Maths, most commonly with combinations of x, exponential functions, logarithms, and trigonometric functions.
The formula
The standard form given in the formula booklet is:
In words: you pick one part of the product to differentiate (u) and the other part to integrate (dv/dx), and the formula trades your original integral for a new one that is, if you have chosen correctly, easier to evaluate.
The one decision that determines everything: choosing u
Almost every mistake in integration by parts traces back to choosing u and dv the wrong way round. The aim is to choose u so that differentiating it makes the expression simpler, and to choose dv so that integrating it stays manageable.
A widely used memory aid for choosing u, in order of priority, is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Whichever function type appears earliest in that list should usually be chosen as u, because logarithmic and inverse trigonometric functions do not simplify when integrated but do simplify when differentiated, while exponential functions do the opposite — they stay essentially unchanged under both operations, which makes them a safe choice for dv.
| Function type in the product | Priority as u | Why |
|---|---|---|
| Logarithmic (e.g. ln x) | Highest | Cannot easily be integrated but differentiates to something simpler (1/x) |
| Inverse trig (e.g. arctan x) | High | Similarly awkward to integrate, simplifies on differentiation |
| Algebraic (e.g. x, x²) | Medium | Differentiates to something simpler each time; integrates without difficulty either way |
| Trigonometric (e.g. sin x, cos x) | Lower | Cycles under repeated differentiation or integration |
| Exponential (e.g. eˣ) | Lowest | Essentially unchanged by differentiation or integration, so usually the best choice for dv |
Worked example 1: x multiplied by an exponential
Find ∫ x eˣ dx.
- Following LIATE, x is algebraic and eˣ is exponential, so choose u = x and dv/dx = eˣ.
- Differentiate u: du/dx = 1.
- Integrate dv/dx: v = eˣ.
- Substitute into the formula: ∫ x eˣ dx = x eˣ − ∫ eˣ · 1 dx.
- Evaluate the remaining integral: ∫ eˣ dx = eˣ.
- Combine and add the constant of integration: x eˣ − eˣ + c.
Notice why the choice worked: differentiating x turned it into 1, removing the algebraic factor entirely and leaving a single, simple integral of eˣ. Had u and dv been swapped, the working would have required integrating x, giving x²/2, and differentiating eˣ, which stays as eˣ — the resulting integral would not have simplified, and the method would stall.
Worked example 2: a logarithm on its own
Find ∫ ln x dx.
This looks like a single function rather than a product, which is exactly the situation where students often assume integration by parts cannot apply. The trick is to treat it as a product with 1: ln x × 1.
- Choose u = ln x (logarithmic, highest priority) and dv/dx = 1.
- Differentiate u: du/dx = 1/x.
- Integrate dv/dx: v = x.
- Substitute: ∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − ∫ 1 dx.
- Evaluate: x ln x − x + c.
Worked example 3: when you need to apply the method twice
Find ∫ x² eˣ dx.
Here x² is algebraic and eˣ is exponential, so u = x², dv/dx = eˣ. Applying the formula once gives ∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx. The remaining integral, ∫ 2x eˣ dx, is itself a product requiring integration by parts — this time with u = 2x, dv/dx = eˣ — which gives 2x eˣ − 2eˣ. Combining both stages: x² eˣ − 2x eˣ + 2eˣ + c.
This pattern — where each application of the formula reduces the power of x by one — is common whenever an algebraic factor has a power of two or more, and it is worth recognising early so that the length of the working does not come as a surprise mid-question.
The self-referencing case
A distinct and commonly examined situation arises with products like eˣ sin x or eˣ cos x, where repeated integration by parts eventually returns the original integral rather than simplifying to zero. The technique here is to apply the formula twice, recognise the original integral reappearing on the right-hand side, and solve for it algebraically as if it were an unknown.
- Let I = ∫ eˣ sin x dx.
- Apply integration by parts once (u = sin x, dv/dx = eˣ): I = eˣ sin x − ∫ eˣ cos x dx.
- Apply integration by parts again to the new integral (u = cos x, dv/dx = eˣ): ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I.
- Substitute back: I = eˣ sin x − (eˣ cos x + I) = eˣ sin x − eˣ cos x − I.
- Solve for I algebraically: 2I = eˣ sin x − eˣ cos x, so I = ½ eˣ (sin x − cos x) + c.
Definite integrals by parts
For a definite integral, the uv term is evaluated between the limits at the same stage as everything else, rather than left until the end: ∫ₐᵇ u (dv/dx) dx = [uv]ₐᵇ − ∫ₐᵇ v (du/dx) dx. A common error is to find the indefinite result first and substitute limits only at the very end, which is not wrong in principle but increases the chance of forgetting to evaluate the boundary term.
Common errors and how to avoid them
- Choosing u and dv the wrong way round, leading to an integral that gets harder rather than simpler — if the new integral is not simpler, stop and check the choice of u before continuing.
- Losing a sign when substituting into uv − ∫ v (du/dx) dx, particularly when u involves a negative or a trigonometric function.
- Forgetting the constant of integration on indefinite integrals.
- In the self-referencing case, switching the choice of u between the two applications, which prevents the original integral from reappearing correctly.
- Not evaluating the boundary (uv) term at the limits on a definite integral, and only applying the limits to the remaining integral.
How to recognise when to use it in an exam
Integration by parts is the right tool when the integrand is a product of two functions of different types (algebraic, exponential, logarithmic, trigonometric) that cannot be simplified by algebraic expansion, and where substitution does not obviously apply — that is, there is no clear 'inner function' whose derivative also appears in the expression. If substitution looks awkward and the expression is clearly a product rather than a composite function, integration by parts is usually the intended method.
The short version
- Integration by parts reverses the product rule: ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx.
- Choose u using LIATE — logarithmic and inverse trig functions first, exponential last.
- A single function like ln x can be integrated by parts by treating it as that function multiplied by 1.
- Powers of x may require applying the method more than once, reducing the power each time.
- Products like eˣ sin x require solving algebraically for the original integral once it reappears.
- On definite integrals, evaluate the uv boundary term at the limits at the same stage as the rest of the working.