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NDA Maths · Indefinite Integration

Integration by Parts

Integration by parts is the product rule run backwards: it trades an integral of a product for a simpler one, choosing which factor to differentiate by the LIATE order.

All Indefinite Integration notesNDA Maths strategy

Why this matters

Only 3 PYQs sit here, all MODERATE — but they are reliable marks once the formula and the LIATE choice are automatic. The NDA favours two shapes: integrating a lone logarithm (treat it as ln x times 1), and a difference of integrals that telescopes once you apply parts. This is also where the eˣ·trig cyclic results from Standard Forms actually come from.

Concept 1 of 3

The By-Parts Formula and LIATE

Intuition

When an integrand is a product of two unlike functions, by parts lets you differentiate one and integrate the other. The art is choosing which is which — LIATE names the priority order for what to differentiate.

Definition

The formula:

∫u dv=uv−∫v du.\int u\,dv = uv - \int v\,du.
Choose uu (the part to differentiate) by LIATE — the earlier in this list, the better a choice for uu:

  • L — Logarithmic (ln⁡x\ln x)
  • I — Inverse trig (tan⁡−1x\tan^{-1}x)
  • A — Algebraic (x2,xx^2, x)
  • T — Trigonometric (sin⁡x\sin x)
  • E — Exponential (exe^x)

Whatever is left becomes dvdv, which you integrate to get vv. A good choice makes ∫v du\int v\,du simpler than the original.

Integration by parts

∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du
  • ufactor you differentiate (pick by LIATE)
  • dvremaining factor, which you integrate to vv

Worked example

Evaluate ∫x ex dx\displaystyle\int x\,e^x\,dx.
  1. LIATE: Algebraic beats Exponential, so u=xu=x (differentiate) and dv=ex dxdv=e^x\,dx (integrate to v=exv=e^x).
  2. Apply the formula: uv−∫v du=xex−∫ex⋅1 dxuv - \int v\,du = x e^x - \int e^x\cdot 1\,dx.
  3. Finish: xex−exx e^x - e^x.
Answer:ex(x−1)+Ce^x(x-1) + C

Choosing u backwards makes it worse

If ∫v du\int v\,du is harder than where you started, you picked uu and dvdv the wrong way round. LIATE almost always points to the right uu — trust it.

Concept 2 of 3

Integrating a Lone Logarithm

Intuition

A logarithm has no elementary integral on its own, so you manufacture a product: write ln x as ln x times 1, then integrate by parts with the 1 as dv. Log laws first turn ln of a power into a constant times ln x.

Definition

The standard result, via parts with u=ln⁡x, dv=dxu=\ln x,\ dv=dx:

∫ln⁡x dx=xln⁡x−x+C.\int \ln x\,dx = x\ln x - x + C.
Use ln⁡(xk)=kln⁡x\ln(x^k) = k\ln x to pull constants out first, so ∫ln⁡(xk) dx=k∫ln⁡x dx=k(xln⁡x−x)+C\int \ln(x^k)\,dx = k\int \ln x\,dx = k(x\ln x - x) + C.

Integral of the logarithm

∫ln⁡x dx=xln⁡x−x+C\int \ln x\,dx = x\ln x - x + C

Worked example

Evaluate ∫ln⁡(x3) dx\displaystyle\int \ln(x^3)\,dx.
  1. Pull the power out: ln⁡(x3)=3ln⁡x\ln(x^3) = 3\ln x, so the integral is 3∫ln⁡x dx3\int \ln x\,dx.
  2. Use the standard result ∫ln⁡x dx=xln⁡x−x\int \ln x\,dx = x\ln x - x.
  3. Multiply by 3.
Answer:3(xln⁡x−x)+C=3xln⁡x−3x+C3(x\ln x - x) + C = 3x\ln x - 3x + C

From the bank · past-year question

Example 2Indefinite IntegrationMODERATE
What is ∫ln⁡(x2) dx\displaystyle\int \ln(x^2)\,dx equal to?

[Q87 · Apr · 2019]

ln x has no naive antiderivative

∫ln⁡x dx\int \ln x\,dx is NOT 1x\dfrac{1}{x} (that is the derivative) and NOT (ln⁡x)22\dfrac{(\ln x)^2}{2}. It is xln⁡x−xx\ln x - x — derive it by parts if you ever forget it.

Concept 3 of 3

Products and Telescoping Cancellations

Intuition

By parts shines on algebraic-times-trig or algebraic-times-exponential products. It also creates clean cancellations: a difference of two awkward integrals can collapse because applying parts to one produces the other.

Definition

Two NDA shapes:

  • Algebraic × trig/exp: e.g. ∫xcos⁡x dx\int x\cos x\,dx with u=xu=x gives xsin⁡x+cos⁡x+Cx\sin x + \cos x + C. Simplify any disguised factor first — eln⁡x=xe^{\ln x} = x turns ∫(eln⁡x+sin⁡x)cos⁡x dx\int (e^{\ln x}+\sin x)\cos x\,dx into ∫xcos⁡x dx+∫sin⁡xcos⁡x dx\int x\cos x\,dx + \int \sin x\cos x\,dx.
  • Telescoping difference: integrals like ∫(ln⁡x)−1 dx−∫(ln⁡x)−2 dx\int (\ln x)^{-1}\,dx - \int (\ln x)^{-2}\,dx collapse because applying parts to one of them throws off exactly the other, leaving a single closed term xln⁡x\dfrac{x}{\ln x}.

The product-rule trade

∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du

Worked example

Evaluate ∫x sin⁡x dx\displaystyle\int x\,\sin x\,dx.
  1. LIATE: u=xu=x (differentiate), dv=sin⁡x dxdv=\sin x\,dx so v=−cos⁡xv=-\cos x.
  2. Apply: uv−∫v du=−xcos⁡x−∫(−cos⁡x) dx=−xcos⁡x+∫cos⁡x dxuv-\int v\,du = -x\cos x - \int(-\cos x)\,dx = -x\cos x + \int \cos x\,dx.
  3. Finish: −xcos⁡x+sin⁡x-x\cos x + \sin x.
Answer:sin⁡x−xcos⁡x+C\sin x - x\cos x + C

From the bank · past-year question

Example 3Indefinite IntegrationMODERATE
∫(ln⁡x)−1 dx−∫(ln⁡x)−2 dx\displaystyle\int (\ln x)^{-1}\,dx - \int (\ln x)^{-2}\,dx is equal to

[Q88 · Sep · 2017]

Simplify disguised factors before applying parts

eln⁡xe^{\ln x} is just xx; eln⁡(tan⁡x)e^{\ln(\tan x)} is just tan⁡x\tan x. Collapse these FIRST — applying by-parts to the disguised form wastes a step and invites errors.
Drill 1 more on products and telescoping cancellations

Summary — formulas & gotchas at a glance

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Mastery check — 1 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Indefinite IntegrationMODERATE
What is ∫(elog⁡x+sin⁡x)cos⁡x dx\int(e^{\log x}+\sin x)\cos x\,dx equal to?

[Q83 · Apr · 2020]

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