Maths › Integration › Integration by substitution and by parts

Integration by substitution and by parts

The chain rule and the product rule each run backwards into a method. Substitution relabels an integral until it looks like something the shelf recognises, and parts trades one integral for a hopefully easier one, with the logarithm's own integral as its most famous conquest.

Year 12-13EDEXCEL 9MA0 8.5

Builds on Integrating standard functions and The product, quotient and chain rules.

IN THIS TOPIC

  • Spot reverse chain rule patterns, f'/f and f' times a power of f, at sight.
  • Run full substitutions, converting integrand, dx and limits together.
  • Integrate by parts, choosing which factor to differentiate, including ∫ln x dx.
  • Apply parts twice where one pass is not enough.

COMMON MISCONCEPTION

∫uv dx = ∫u dx × ∫v dx.

Reverse chain patterns

The chain rule leaves fingerprints. When an integrand contains a function and its derivative standing alongside, the integral undoes a chain in one step, and three patterns cover most of what the papers set.

∫f′(x)f(x)dx=ln|f(x)|+c\text{∫} \frac{f'(x)}{f(x)} \, dx = \text{ln}|f(x)| + cNOT IN THE BOOKLET — LEARN IT

The modulus signs are doing work: the result holds on any interval where f keeps the same sign, and says nothing across a point where f is zero. On a definite integral, check that the limits do not straddle a root of f before writing it down.

The other two are f' times a power of f, which reverses to the next power up, and the famous special case ∫tan x dx = −ln|cos x| + c, which is the first pattern again with sin and cos in the roles.

Three reverse chain rule patterns: 2x over x squared plus 1 gives log of the bottom, sine cubed x cos x gives a power, and tan x gives minus log of cos x2x/(x² + 1)ln(x² + 1) + cspot f’/fsin³x cos x¼ sin⁴x + cspot f’ fⁿtan x−ln|cos x| + cf’/f in itthe inside function's derivative, standing alongside, is the tell
FIG. 1The three tells: derivative-over-function makes a log, derivative-times-power reverses the chain, and tan x is the first pattern again.

WORKED EXAMPLE

A full substitution, limits and all

Use the substitution u = 2x + 1 to find ∫x√(2x + 1) dx.

From u = 2x + 1: x = (u − 1)/2 and dx = du/2, so the integral becomes ∫((u − 1)/2)√u (du/2) = ¼∫(u3/2 − u1/2) du.

Integrating: ¼((2/5)u5/2 − (2/3)u3/2) = u5/2/10 − u3/2/6.

Back in x: (2x + 1)5/2/10 − (2x + 1)3/2/6 + c.

Three things convert or nothing does. The integrand, the dx, and on a definite integral the limits as well. Omitting the dx factor of ½ is the most frequent error in this method.

GUIDED PRACTICE

Spot, do not substitute

Find ∫sin3 x cos x dx by inspection, before opening the working.

Show the working

cos x is the derivative of sin x, standing right beside its powers. That is the f' fⁿ pattern.

Reverse the chain: ¼ sin4 x + c, and differentiating back confirms it in one line.

A formal substitution u = sin x reaches the same place. Inspection is the same route, travelled lighter.

Integration by parts

The product rule reverses into parts, printed in the booklet as

∫udvdxdx=uv−∫vdudxdx\text{∫} u \frac{dv}{dx} \, dx = uv − \text{∫} v \frac{du}{dx} \, dxIN THE FORMULAE BOOKLET

which is a trade. One factor gets differentiated, the other gets integrated, and the original integral becomes a new one that had better be easier. Choose to differentiate the factor that simplifies, x going to 1, ln x going to 1/x. Factor-by-factor integration fails by counter example: integrate x and ex separately and you get (x2/2)ex, whose derivative is nowhere near xex.

WORKED EXAMPLE

The standard parts question

Find ∫xex dx.

Differentiate the x, which simplifies to 1, and integrate the ex, which costs nothing. Parts gives xex − ∫ex dx.

∫xex dx = (x − 1)ex + c.

The trade turned a product into the easiest integral in the course. Choose the roles the other way round and the new integral is harder than the old one, which is your diagnostic for a wrong choice.

Integration by parts on the log: writing ln x as 1 times ln x turns the integral into x ln x minus the integral of 1, which is x ln x minus x plus c∫ ln x dx = ∫ (1 × ln x) dxintegrate the 1 → xdifferentiate ln x → 1/x= x ln x − ∫ x × (1/x) dx = x ln x − ∫ 1 dx= x ln x − x + cthe specification asks for this one by name
FIG. 2The specification's named target: ∫ln x dx by parts, writing ln x as 1 × ln x. Differentiate the log, integrate the 1, and the result is x ln x − x + c.

INDEPENDENT PRACTICE

The logarithm's own integral

Find ∫ln x dx.

Show the working

Write the integrand as 1 × ln x, then differentiate the ln x and integrate the 1.

Parts gives x ln x − ∫x × (1/x) dx = x ln x − ∫1 dx.

∫ln x dx = x ln x − x + c, and differentiating back, x/x + ln x − 1 = ln x, seals it.

That invisible factor of 1 is what makes the whole thing go. The specification names this integral, so learn the answer as well as the method.

When once is not enough

The specification allows more than one application of parts, and ∫x2ex dx is the standard place it shows up. Differentiate the x2 and it drops to 2x, which is progress but not an answer. So run parts again on what is left. Each pass knocks the power down by one, and a polynomial of degree n needs n passes.

WORKED EXAMPLE

Parts, twice

Find ∫x2ex dx.

First pass, differentiating x2: x2ex − ∫2xex dx.

The remaining integral is the worked example above, doubled: ∫2xex dx = 2(x − 1)ex.

So ∫x2ex dx = x2ex − 2(x − 1)ex + c = (x2 − 2x + 2)ex + c.

Watch the sign on the second pass. You are subtracting a bracket that itself contains a subtraction, and that is where nearly every lost mark in double parts comes from.

ASSESSMENT FOCUS

  • Scan for a function sitting beside its own derivative before reaching for substitution. The patterns are faster and carry the same marks.
  • In a substitution, convert integrand, dx and limits together, and say that you have. Limits left in terms of x are the usual reason a definite integral comes out wrong.
  • In parts, differentiate the factor that simplifies. If the new integral is harder than the old one, the roles were the wrong way round.
  • ∫tan x dx = −ln|cos x| + c is quotable, and its one-line derivation from f'/f is worth knowing cold.
  • A second application of parts needs the whole bracket subtracted. Write it out; do not distribute the minus sign in your head.
  • Check every antiderivative by differentiating. Substitution and parts both reverse cleanly and instantly.

CHECK YOURSELF

Find the exact value of ∫01 xex dx.

Show a hint

Parts, then substitute both limits into (x − 1)ex.

Show the answer

Parts gives the antiderivative (x − 1)ex.

Evaluating: (1 − 1)e − (0 − 1) × 1 = 0 + 1 = 1.

An area of exactly 1 under xex from 0 to 1, and every step of it exact. The bracket at the lower limit is where the sign errors hide.

A function beside its own derivative is a reversed chain; relabel or just read it off.

Parts trades one integral for another: differentiate what simplifies, and run it twice when once will not do.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

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Question workbook

PDF · 32 KB · screen edition · navy

Answer book

PDF · 59 KB · fully worked · navy
8 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the integration by substitution and by parts questions page.

CHECK YOUR PROGRESS

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  • Spot reverse chain rule patterns, f'/f and f' times a power of f, at sight.
  • Run full substitutions, converting integrand, dx and limits together.
  • Integrate by parts, choosing which factor to differentiate, including ∫ln x dx.
  • Apply parts twice where one pass is not enough.

Open the full revision checklist to see every objective in the course in one place.