Integration by Parts

This lesson covers integration by parts, a technique for integrating products of two functions. It is the reverse of the product rule for differentiation and is essential for A-Level Mathematics.

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The Formula

Integration by parts states:

∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx

Or in the shorter notation:

∫ u dv = uv − ∫ v du

This formula converts one integral into another that is hopefully simpler.

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Derivation from the Product Rule

The product rule states:

d/dx [uv] = u(dv/dx) + v(du/dx)

Integrating both sides:

uv = ∫ u(dv/dx) dx + ∫ v(du/dx) dx

Rearranging:

∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx

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Choosing u and dv/dx

The key to successful integration by parts is choosing which factor is u and which is dv/dx. The goal is to make ∫ v(du/dx) dx simpler than the original integral.

A useful guideline (sometimes called LIATE) is to let u be the function that comes first in this list:

1. Logarithmic functions (ln x)
2. Inverse trigonometric functions (arcsin x, arctan x)
3. Algebraic functions (x², x, polynomials)
4. Trigonometric functions (sin x, cos x)
5. Exponential functions (eˣ)

The remaining factor is dv/dx.

Key Point: Let u be the function that becomes simpler when differentiated. Let dv/dx be the function that you can integrate.

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Example 1: ∫ x eˣ dx

Choose: u = x (algebraic — differentiates to something simpler), dv/dx = eˣ.

Then: du/dx = 1, v = eˣ.

∫ x eˣ dx = x eˣ − ∫ eˣ × 1 dx
= x eˣ − eˣ + C
= eˣ(x − 1) + C

Check: d/dx [eˣ(x − 1)] = eˣ(x − 1) + eˣ = eˣ × x − eˣ + eˣ = x eˣ ✓

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Example 2: ∫ x cos x dx

Choose: u = x, dv/dx = cos x.

Then: du/dx = 1, v = sin x.

∫ x cos x dx = x sin x − ∫ sin x dx
= x sin x + cos x + C

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Example 3: ∫ x² eˣ dx (Repeated Application)

Sometimes you need to apply integration by parts twice (or more).

First application:

u = x², dv/dx = eˣ → du/dx = 2x, v = eˣ.

∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx

Second application (on ∫ 2x eˣ dx):

u = 2x, dv/dx = eˣ → du/dx = 2, v = eˣ.

∫ 2x eˣ dx = 2x eˣ − ∫ 2eˣ dx = 2x eˣ − 2eˣ + C

Combining:

∫ x² eˣ dx = x² eˣ − (2x eˣ − 2eˣ) + C
= x² eˣ − 2x eˣ + 2eˣ + C
= eˣ(x² − 2x + 2) + C

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Example 4: ∫ ln x dx

This is a classic integral that uses a clever trick: treat ln x as a product of ln x and 1.

Choose: u = ln x, dv/dx = 1.

Then: du/dx = 1/x, v = x.

∫ ln x dx = x ln x − ∫ x × (1/x) dx
= x ln x − ∫ 1 dx
= x ln x − x + C

This result is important and worth memorising:

∫ ln x dx = x ln x − x + C

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Example 5: ∫ x ln x dx

Choose: u = ln x (LIATE — logs first), dv/dx = x.

Then: du/dx = 1/x, v = x²/2.

∫ x ln x dx = (x²/2) ln x − ∫ (x²/2)(1/x) dx
= (x²/2) ln x − (1/2) ∫ x dx
= (x²/2) ln x − x²/4 + C

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Definite Integrals by Parts

For definite integrals, the formula becomes:

∫(from a to b) u (dv/dx) dx = [uv](from a to b) − ∫(from a to b) v (du/dx) dx

Example 6

Evaluate ∫₁ᵉ ln x dx.

From Example 4, ∫ ln x dx = x ln x − x.

∫₁ᵉ ln x dx = [x ln x − x]₁ᵉ
= (e ln e − e) − (1 ln 1 − 1)
= (e − e) − (0 − 1)
= 0 + 1
= 1

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Example 7: The "Cycling" Trick — ∫ eˣ sin x dx

When both functions regenerate upon repeated differentiation/integration, the integral "cycles back" to itself.

First application:

u = eˣ, dv/dx = sin x → du/dx = eˣ, v = −cos x.

I = ∫ eˣ sin x dx = −eˣ cos x + ∫ eˣ cos x dx

Second application (on ∫ eˣ cos x dx):

u = eˣ, dv/dx = cos x → du/dx = eˣ, v = sin x.

∫ eˣ cos x dx = eˣ sin x − ∫ eˣ sin x dx = eˣ sin x − I

Substitute back:

I = −eˣ cos x + eˣ sin x − I
2I = eˣ(sin x − cos x)
I = (eˣ/2)(sin x − cos x) + C

Exam Tip: When integrating eˣ sin x or eˣ cos x by parts, after two applications you get the original integral on the right-hand side. Call the original integral I, rearrange to solve for I. This is a favourite AQA question.

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Example 8: ∫ x² sin x dx

This requires integration by parts applied twice.

First: u = x², dv/dx = sin x → du/dx = 2x, v = −cos x.

∫ x² sin x dx = −x² cos x + ∫ 2x cos x dx

Second: u = 2x, dv/dx = cos x → du/dx = 2, v = sin x.

∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx = 2x sin x + 2 cos x + C

Combine:

∫ x² sin x dx = −x² cos x + 2x sin x + 2 cos x + C

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Common Mistakes

1. Choosing u and dv/dx badly — if your choice makes the integral harder, swap them.
2. Sign errors — be very careful with the minus sign in uv − ∫ v du.
3. Forgetting + C in indefinite integrals.
4. Not applying parts enough times — some integrals need two (or occasionally three) applications.
5. Cycling integrals — recognise when the original integral reappears and solve algebraically for I.

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Summary

• Integration by parts: ∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx.
• Use LIATE to guide your choice of u.
• For ∫ ln x dx: let u = ln x, dv/dx = 1 → result is x ln x − x + C.
• Sometimes you must apply parts twice (e.g. x² eˣ, x² sin x).
• For cycling integrals (eˣ sin x, eˣ cos x), apply twice and solve for I.
• For definite integrals, evaluate [uv] at the limits as well.

Exam Tip: Integration by parts questions are worth several marks. Show every step: state your choice of u and dv/dx, write down du/dx and v, substitute into the formula, and simplify. If you need to apply parts twice, make it clear where the second application begins. Always check your answer by differentiating if time allows.

A-Level Deep Dive: Integration by Parts

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section H (Integration), sub-strand H8 — Year 2 content covers using integration by parts; integrate using partial fractions; integrate by recognition or by simple substitution; understand integration as the limit of a sum (refer to the official specification document for exact wording). Integration by parts is examined principally on Paper 2 but synoptic reach extends into Paper 3 (Section O — Differential equations) where by-parts integration appears within solution techniques. The AQA formula booklet does list $\int u \dfrac{dv}{dx}\,dx = uv - \int v \dfrac{du}{dx}\,dx$∫udxdv​dx=uv−∫vdxdu​dx, so the formula itself need not be memorised — but the strategy of choosing $u$u and $dv$dv is a candidate decision and is never prompted in the question stem.

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Worked example with full mark scheme

Question (8 marks): Find $\int x^2 e^x\,dx$∫x2exdx.

Solution with mark scheme:

Step 1 — choose $u$u and $dv$dv on the first application.

By LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential), the algebraic factor $x^2$x2 ranks higher than the exponential $e^x$ex, so set $u = x^2$u=x2 and $dv = e^x\,dx$dv=exdx.

Then $du = 2x\,dx$du=2xdx and $v = e^x$v=ex.

M1 — correct identification of $u$u and $dv$dv consistent with LIATE; equivalently, choosing $u$u such that $du$du simplifies the integrand. Common error: setting $u = e^x$u=ex and $dv = x^2\,dx$dv=x2dx, which produces a more complex integral on the right-hand side and stalls the question.

Step 2 — apply the formula.

$\int x^2 e^x\,dx = x^2 e^x - \int e^x \cdot 2x\,dx = x^2 e^x - 2\int x e^x\,dx$∫x2exdx=x2ex−∫ex⋅2xdx=x2ex−2∫xexdx

M1 — correct substitution into $\int u\,dv = uv - \int v\,du$∫udv=uv−∫vdu.

A1 — correct intermediate expression $x^2 e^x - 2\int x e^x\,dx$x2ex−2∫xexdx.

Step 3 — apply integration by parts a second time to $\int x e^x\,dx$∫xexdx.

Set $u = x$u=x, $dv = e^x\,dx$dv=exdx, so $du = dx$du=dx and $v = e^x$v=ex.

$\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C_1$∫xexdx=xex−∫exdx=xex−ex+C1​

M1 — correct second application of integration by parts on the residual integral. Many candidates panic at the reappearance of an integral and try a substitution; the principled move is to apply by-parts again because the algebraic factor has reduced from $x^2$x2 to $x$x (and will reduce to a constant on a third pass — except a constant integrates trivially).

A1 — correct evaluation $xe^x - e^x$xex−ex.

Step 4 — assemble the final answer.

$\int x^2 e^x\,dx = x^2 e^x - 2(x e^x - e^x) + C = x^2 e^x - 2x e^x + 2 e^x + C$∫x2exdx=x2ex−2(xex−ex)+C=x2ex−2xex+2ex+C

Factorising: $= e^x(x^2 - 2x + 2) + C$=ex(x2−2x+2)+C.

M1 — correct combination, with sign on the $-2(\ldots)$−2(…) bracket distributed properly. The most common slip is sign error: writing $+2(xe^x - e^x)$+2(xex−ex) instead of $-2(xe^x - e^x)$−2(xex−ex).

A1 — correct final expression including the constant of integration $+C$+C. Omitting $+C$+C is an automatic loss of the final A1 on every indefinite integration question, every paper, every year.

A1 — factorised or simplified form (rewards presentation).

Total: 8 marks (M4 A4).

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Specimen question modelled on AQA 7357 Paper 2 format

Question (6 marks): Use integration by parts to find the exact value of $\int_1^e x \ln x\,dx$∫1e​xlnxdx.

Mark scheme decomposition by AO:

• M1 (AO1.1a) — choosing $u = \ln x$u=lnx, $dv = x\,dx$dv=xdx, giving $du = \dfrac{1}{x}dx$du=x1​dx and $v = \dfrac{x^2}{2}$v=2x2​. LIATE places L above A, so $\ln x$lnx becomes $u$u — and crucially, $du = (1/x)dx$du=(1/x)dx kills the logarithm in the residual integral.
• M1 (AO1.1b) — correct substitution: $\int x \ln x\,dx = \dfrac{x^2}{2}\ln x - \int \dfrac{x^2}{2} \cdot \dfrac{1}{x}\,dx = \dfrac{x^2}{2}\ln x - \int \dfrac{x}{2}\,dx$∫xlnxdx=2x2​lnx−∫2x2​⋅x1​dx=2x2​lnx−∫2x​dx.
• A1 (AO1.1b) — correct antiderivative $\dfrac{x^2}{2}\ln x - \dfrac{x^2}{4}$2x2​lnx−4x2​.
• M1 (AO1.1b) — correct substitution of bounds: at $x = e$x=e, $\dfrac{e^2}{2} \cdot 1 - \dfrac{e^2}{4} = \dfrac{e^2}{4}$2e2​⋅1−4e2​=4e2​; at $x = 1$x=1, $\dfrac{1}{2} \cdot 0 - \dfrac{1}{4} = -\dfrac{1}{4}$21​⋅0−41​=−41​.
• A1 (AO1.1b) — exact answer $\dfrac{e^2}{4} - \left(-\dfrac{1}{4}\right) = \dfrac{e^2 + 1}{4}$4e2​−(−41​)=4e2+1​.
• A1 (AO2.5) — answer presented as a single simplified fraction in exact form (not a decimal).

Total: 6 marks split AO1 = 5, AO2 = 1. AQA reserves the AO2 mark for "exact form" presentation discipline — any decimal approximation forfeits it.

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Synoptic links

Connects to:

1. Section G — Differentiation (product rule, reverse direction): integration by parts is literally the product rule $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$dxd​(uv)=udxdv​+vdxdu​ rearranged and integrated. Recognising $\int u\,dv = uv - \int v\,du$∫udv=uv−∫vdu as "anti-product-rule" demystifies why the formula has the structure it does.

2. LIATE choice and AO2 reasoning: the choice of $u$u is not algorithmic — it is a reasoning skill. AQA increasingly tests this with questions where a poor choice produces a harder integral. The LIATE heuristic ranks function classes by how well $u$u simplifies under differentiation: Logarithmic and Inverse-trig functions simplify dramatically ($\ln x \to 1/x$lnx→1/x); Algebraic factors reduce in degree ($x^n \to nx^{n-1}$xn→nxn−1); Trigonometric and Exponential functions cycle without simplifying.

3. Reduction formulae (Year 2 / undergraduate): repeated integration by parts on $\int x^n e^x\,dx$∫xnexdx produces the recurrence $I_n = x^n e^x - n I_{n-1}$In​=xnex−nIn−1​. AQA does not formally examine reduction formulae at A-Level, but recognising the pattern in $\int x^2 e^x\,dx$∫x2exdx (two passes) and $\int x^3 e^x\,dx$∫x3exdx (three passes) is exactly the synoptic insight that distinguishes A* candidates.

4. Combined with substitution: harder Paper 2 integrals require both substitution and by-parts. Example: $\int \sin(\sqrt{x})\,dx$∫sin(x​)dx needs substitution $u = \sqrt{x}$u=x​ first to convert it into $\int 2u \sin u\,du$∫2usinudu, which then requires by-parts. The strategic order — substitute first, then by-parts — is rarely signposted in the question.

5. Section O — Differential equations: solving $\dfrac{dy}{dx} + P(x)y = Q(x)$dxdy​+P(x)y=Q(x) via the integrating-factor method produces integrals of the form $\int Q(x) \mu(x)\,dx$∫Q(x)μ(x)dx where $\mu$μ is exponential. When $Q(x)$Q(x) is algebraic, the resulting integral demands by-parts. This is the most common cross-section question on Paper 2.

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Mark-scheme literacy

By-parts questions on AQA 7357 split AO marks predictably:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Correctly identifying $u$u, $dv$dv, computing $du$du, $v$v, applying the formula, evaluating residual integral
AO2 (reasoning / interpretation)	20–30%	Justifying LIATE choice, recognising when a second pass is needed, presenting answer in exact form
AO3 (problem-solving)	10–15%	Combining by-parts with substitution; recognising hidden by-parts structures (e.g. $\int \ln x\,dx$∫lnxdx as $\int 1 \cdot \ln x\,dx$∫1⋅lnxdx)