Integration by parts — derivation from product rule, LIATE mnemonic: Question bank

True or false — justify

Integration by parts is a brand-new rule unrelated to differentiation.

False. It is literally the product rule integrated on both sides — nothing new is assumed, it is derived.

The formula

\int u d v = uv - \int v d u

always makes an integral easier.

False. It only helps if the new integral

\int v d u

is simpler than the original; a bad choice of

u

and

d v

can make it harder or send you in circles.

You can apply integration by parts to any product of two functions.

True in principle, useful only sometimes. The rule is always valid, but it's only worth doing when one factor simplifies on differentiating and the other is easy to integrate.

\int ln x d x

cannot be done by parts because there is only one function.

False. Write

ln x = ln x \cdot 1

; the hidden factor

1

becomes

d v

, so it is a product after all.

The "

+ C

" is optional in integration by parts because the

uv

term absorbs it.

False.

uv

is a specific function, not an arbitrary constant; every indefinite integral still needs

+ C

If doing parts once leaves another product, the method has failed.

False. Repeating parts is expected — each pass on a polynomial drops its degree by one until it vanishes (e.g.

\int x^{2} cos x d x

needs two passes).

The choice of

u

and

d v

can change the final answer.

False. A valid choice may make the work harder, but any correct route gives the same antiderivative (up to

+ C

); a bad choice just costs effort.

For a definite integral, the

uv

term must also be evaluated at the limits.

True. On definite integrals the boundary term becomes

[uv]_{a}^{b}

; forgetting to plug in the limits there is a classic error.

Substitution and integration by parts are interchangeable — pick either freely.

False. Try substitution first (it's simpler); parts is for products that no single substitution untangles.

The "loop" trick (

\int e^{x} sin x d x

) works because the integral literally disappears.

False. The original integral reappears, and you solve for it algebraically like an unknown — it doesn't vanish, it gets isolated.

Spot the error

"For

\int x e^{x} d x

I set

u = e^{x}

because

e^{x}

integrates so nicely."

The rule wants

u

to simplify when differentiated. With

u = e^{x}

d u = e^{x} d x

never simplifies and

\int v d u

gets worse; LIATE says A before E, so

u = x

"I split

\int x e^{x} d x

into pieces

x

and

e^{x}

, and took

d v = e^{x}

d v

must carry the

d x

d v = e^{x} d x

. Since

v = \int d v

, the

d x

marks exactly what is being integrated.

\int ln x d x = x ln x - \int x \cdot x d x

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

, not

x d x

. The integrand becomes

x \cdot \frac{1}{x} = 1

, giving

x ln x - x + C

"After one pass I got

x^{2} sin x - \int 2 x sin x d x

, and I dropped the minus on the second pass."

The leading

-

\int v d u

carries through every pass; losing it flips the sign of the whole tail. Rewrite the boxed formula fresh each pass to keep it.

\int e^{x} sin x d x

looped back to itself, so it has no answer."

The loop is the feature:

I = e^{x} sin x - (e^{x} cos x + I)

gives

2 I = e^{x} (sin x - cos x)

, so

I = \frac{1}{2} e^{x} (sin x - cos x) + C

"I chose

u = x

d v = cos x d x

, but then wrote

v = - sin x

Integrating

cos x

gives

v = + sin x

; the sign belongs to differentiating

cos x

, not integrating it. Mixing up derivative/integral signs of trig is the most common slip.

"For a definite integral I applied the formula but only put limits on the

\int v d u

term."

Both the boundary term and the remaining integral are evaluated at the limits:

\int_{a}^{b} u d v = [uv]_{a}^{b} - \int_{a}^{b} v d u

Why questions

Why does LIATE put Logarithmic and Inverse-trig first as

u

They are awful to integrate but have clean, simplifying derivatives, so they are ideal as the factor you differentiate.

Why does LIATE put Exponential last?

e^{x}

integrates to itself effortlessly, so it makes an ideal

d v

— the easy-to-integrate factor.

Why does the derivation start from the product rule rather than inventing something?

Integration undoes differentiation, so any differentiation identity, when integrated, yields an integration rule — reusing a trusted identity beats guessing.

Why is

\int \frac{d}{d x} (uv) d x = uv

(no integral sign left)?

The Fundamental Theorem of Calculus says integrating a derivative returns the original function.

Why do we want

u

to vanish eventually for polynomials?

Each pass differentiates the polynomial, lowering its degree; once it hits a constant then

0

, the remaining integral is trivial and the recursion stops.

Why must

d v

be something you can actually integrate?

Because you need

v = \int d v

to even write the formula; if you can't find

v

, you can't apply parts with that choice.

Why is the "hidden product"

ln x = ln x \cdot 1

a legitimate move?

Multiplying by

1

changes nothing about the function but exposes an integrable factor (

d v = 1 d x

) so parts becomes applicable.

Why can repeated parts sometimes be organised with tabular integration?

When one factor differentiates to zero after finitely many steps, the alternating

uv

terms follow a fixed pattern that a table bookkeeps automatically.

Edge cases

What if you differentiate a polynomial factor to zero — does the process end?

Yes; once

d u = 0

the trailing

\int v d u = 0

, so no integral remains and you simply read off the answer.

What happens if you keep applying parts and never simplify (e.g. wrong choice on

\int e^{x} sin x

You loop forever unless you recognise the returning integral and solve for it algebraically, or unless you swap your

u

d v

choice.

What if both factors get harder when you differentiate them (say two trig functions)?

LIATE still gives a rule (T before... whichever), but often a trig identity or substitution is smarter than parts — parts isn't obligatory.

Does integration by parts work when

u

v

isn't differentiable everywhere?

The derivation assumes both are differentiable; at points where a factor isn't smooth, the formula's justification breaks and you must handle those points separately.

What if the "

v

" you pick has its own

+ C

— does it matter?

Any antiderivative

v

works; the arbitrary constant cancels out of the final result, so you conventionally take the simplest

v

(constant

= 0

\int u d v = uv - \int v d u

ever worse than just guessing an antiderivative?

Yes — for simple integrals or ones a single substitution solves, parts adds pointless work; it's a tool for stubborn products, not a first resort.

Connections

Product rule (differentiation) — the identity every trap traces back to.
Fundamental Theorem of Calculus — why the derivative integrates away cleanly.
Integration by substitution — the technique to try before reaching for parts.
Reduction formulae — repeated parts, packaged as a recursion.
Tabular integration (DI method) — bookkeeping for many-pass parts.
Definite integrals — where the boundary term $[uv]_{a}^{b}$ bites.