H5PureStretch

Integration by Parts

Q: Why is ln x always chosen as u?

Because there is no direct rule (yet) to integrate x on its own — but its derivative 1x is simple, so it must be the differentiated part, never the integrated part.

Substitution handles a function multiplied by its own derivative — but what about $\int x sin x d x$ , where the two factors are completely unrelated? Integration by parts reverses the product rule instead, turning one hard integral into an easier one.

34 min Video by TLMaths Integration

The big picture

Integration by parts is the mirror image of the product rule for differentiation. When an integral is a product of two genuinely different types of function (a polynomial times a trig function, or times $e^{x}$ , or times $ln x$ ), this technique transfers the differentiation from one factor onto the other — repeatedly, if necessary — until what remains is something you can integrate directly. Choosing which factor to differentiate and which to integrate is the entire skill.

A-Level Maths H5-26: Further Integration: Introducing Integration by Parts & LATEWatch the walkthrough, then work through the notes below.

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What you'll be able to do

State and apply the integration by parts formula
Use the LATE priority (Logs, Algebraic, Trig, Exponential) to choose u and dv
Apply integration by parts to definite integrals
Recognise when integration by parts must be applied more than once

The integration by parts formula

Reversing the product rule for differentiation gives a formula that swaps which factor is differentiated: you differentiate one part ( $u$ ) and integrate the other ( $\frac{d v}{d x}$ ), and the resulting integral is (hopefully) easier than the original.

\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x

Choose u to differentiate, dv/dx to integrate.

1Let

u = x

(differentiates to something simpler) and

\frac{d v}{d x} = cos x

\frac{d u}{d x} = 1

; integrating

\frac{d v}{d x}

gives

v = sin x

3Apply the formula:

\int x cos x d x = x sin x - \int sin x d x

4Integrate the remaining piece:

\int sin x d x = - cos x

Answer

x sin x + cos x + c

Tip — Choosing $u = x$ was deliberate: differentiating it gives just $1$ , making the new integral simpler. Choosing $u = cos x$ instead would have made things WORSE, not better.

Choosing u: the LATE priority

A useful memory aid for choosing $u$ is $LATE$ : prefer, in order, $L$ ogarithmic, $A$ lgebraic (powers of $x$ ), $T$ rigonometric, $E$ xponential. The earlier a function type appears in this list, the more likely it should be $u$ (since differentiating it tends to simplify things, or it cannot easily be integrated at all, as with $ln x$ ).

1By LATE, logarithmic beats algebraic: let

u = ln x

\frac{d v}{d x} = x

\frac{d u}{d x} = \frac{1}{x}

; integrating gives

v = \frac{x ^{2}}{2}

3Apply the formula:

\int x ln x d x = \frac{x ^{2}}{2} ln x - \int \frac{x ^{2}}{2} \times \frac{1}{x} d x = \frac{x ^{2}}{2} ln x - \int \frac{x}{2} d x

4Integrate the remaining piece:

\frac{x ^{2}}{4}

Answer

\frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} + c

Why this works

$ln x$ cannot be integrated using any rule met so far, so it MUST be chosen as $u$ (to be differentiated, not integrated) whenever it appears in a product.

Repeated application

Sometimes the new integral produced still contains a product requiring the SAME technique again — apply integration by parts a second time, keeping the same choice pattern for $u$ (e.g. still the algebraic factor) consistently.

1Let

u = x^{2}

\frac{d v}{d x} = e^{x}

, so

\frac{d u}{d x} = 2 x

v = e^{x}

2Apply once:

\int x^{2} e^{x} d x = x^{2} e^{x} - \int 2 x e^{x} d x

3The remaining integral

\int 2 x e^{x} d x

still needs parts: let

u = 2 x

\frac{d v}{d x} = e^{x}

, giving

\int 2 x e^{x} d x = 2 x e^{x} - \int 2 e^{x} d x = 2 x e^{x} - 2 e^{x}

4Combine:

x^{2} e^{x} - (2 x e^{x} - 2 e^{x}) = x^{2} e^{x} - 2 x e^{x} + 2 e^{x}

Answer

x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + c

Think like an examiner

Use LATE to decide $u$ : Logarithmic, Algebraic, Trigonometric, Exponential, in that priority order.
Write out $u$ , $\frac{d u}{d x}$ , $\frac{d v}{d x}$ and $v$ clearly before substituting into the formula — this structure earns method marks even if the final algebra goes wrong.
If the new integral is still a product needing parts again, repeat the SAME choice logic consistently.

Common misconceptions

Choosing u and dv/dx the wrong way round, making the new integral harder.

Use LATE: logarithms and algebraic (polynomial) terms are usually chosen as

u

, since they simplify (or cannot be integrated) when differentiated.

Forgetting the minus sign in the formula

uv - \int v \frac{d u}{d x} d x

The formula subtracts the new integral — dropping this sign is a very common algebra slip.

Stopping after one application when the resulting integral is still a product.

If the new integral still needs integration by parts, apply it again rather than giving up or guessing.

Integration by parts

\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x

The integration by parts formula.

LATE priority for u : Log, Algebraic, Trig, Exp

Choosing which factor to differentiate.

Stretch yourself

Evaluate $\int_{0}^{1} x e^{2 x} d x$ using integration by parts, giving an exact answer in terms of $e$ .

Hint — Let $u = x$ , $\frac{d v}{d x} = e^{2 x}$ ; remember integrating $e^{2 x}$ needs the reverse chain rule, giving $v = \frac{1}{2} e^{2 x}$ .

Questions students ask

How do I decide which part is u and which is dv/dx?

Use LATE: prefer Logarithmic, then Algebraic, then Trigonometric, then Exponential as

u

. Whichever type appears earliest in that list should usually be

u

What if the new integral is still hard?

If it is still a product of unrelated function types, apply integration by parts again — keeping the same consistent choice pattern (e.g. always choosing the algebraic/polynomial part as

u

Why is ln x always chosen as u?

Because there is no direct rule (yet) to integrate

ln x

on its own — but its derivative

\frac{1}{x}

is simple, so it must be the differentiated part, never the integrated part.

Key takeaways

Integration by parts reverses the product rule: $\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x$ .
Choose $u$ using LATE priority: Logarithmic, Algebraic, Trigonometric, Exponential.
Apply repeatedly if the resulting integral is still a product needing the technique.
Works for both indefinite and definite integrals (evaluate the uv term at the limits too).

How this fits the course

Build on

Differential Equations

Leads to

Differential Equations

Test yourself

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