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Module 15 · L03 of 16 ~45 min ⚡ +90 XP available

Integration by Parts I

When the integrand is a product — $x \sin x$, $x e^x$, $\ln x$ — substitution fails. Integration by parts is the reverse of the product rule and the standard tool for these. This lesson sets up the formula $\int u\,dv = uv - \int v\,du$, shows how the LIATE mnemonic chooses $u$ and $dv$, and walks through three benchmark examples — including the famous trick for $\int \ln x\,dx$.

Today's hook — Try to evaluate $\int x \sin x \, dx$ by substitution. Let $u = \sin x$? Let $u = x$? Whatever you try, you'll be stuck with another product to integrate. Now compute $\dfrac{d}{dx}(x \cos x)$ and notice what appears. That observation is integration by parts in disguise.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Write out the product rule for differentiation: $\dfrac{d}{dx}(uv) = \;?$. Then integrate both sides with respect to $x$ and rearrange to isolate $\int u\,dv$. What formula falls out?

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The two moves for parts integration

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Every parts question rewards two habits: split the integrand into $u$ and $dv$ (one to differentiate, one to integrate), then apply $\int u\,dv = uv - \int v\,du$ and hope the new integral is easier. Choosing $u$ correctly is the entire art of the technique — and that's where LIATE helps.

The split-apply-simplify reading: (1) label one factor $u$ (it will be differentiated) and the rest $dv$ (it will be integrated), (2) compute $du$ and $v$, (3) substitute into $\int u\,dv = uv - \int v\,du$ and evaluate the new integral.

LIATE: Log · Inverse trig · Algebraic · Trig · Exponential — pick $u$ from the highest class present.

$\int u\,dv \;=\; uv \;-\; \int v\,du$

$u$ should get simpler when differentiated

Pick $u$ so $du$ is no worse than $u$. Polynomials drop a degree; logs become $\tfrac{1}{x}$. If $du$ is uglier than $u$, swap your choice.

$dv$ must be integrable

Whatever you call $dv$, you need $v = \int dv$ in closed form. Exponentials, $\sin x$, $\cos x$, and powers of $x$ all qualify. $\ln x$ does not.

Trick: $dv = dx$

For $\int \ln x\,dx$ or $\int \tan^{-1} x\,dx$ there is no product. Force one by writing $u = \ln x$ and $dv = dx$. Then $v = x$ and the new integral is easy.

What you'll master

Know

Key facts

$\int u\,dv = uv - \int v\,du$, the integration by parts formula
The formula is the product rule for differentiation, integrated
LIATE priority for choosing $u$: Log, Inverse trig, Algebraic, Trig, Exponential
Adding $+C$ at the end is mandatory for indefinite integrals

Understand

Concepts

Why parts converts $\int u\,dv$ into a different (hopefully easier) integral
Why LIATE works — the higher class produces a simpler $du$
Why $dv = dx$ unlocks single-function integrands like $\ln x$

Can do

Skills

Apply the IBP formula to integrate $x \sin x$, $x e^x$, $x \cos x$, $x \ln x$ etc.
Evaluate $\int \ln x\,dx$ and similar single-function integrals using $dv = dx$
Use LIATE to make the right choice of $u$ on first attempt

Key terms

Integration by parts (IBP)The integral identity $\int u\,dv = uv - \int v\,du$, obtained by integrating the product rule.

$u$ (the chosen factor)The factor of the integrand we differentiate. Choose so $du$ is simpler than $u$.

$dv$ (the remainder)Everything else, including $dx$. We integrate to find $v$; it must be integrable in closed form.

LIATEMnemonic ranking — Log, Inverse trig, Algebraic, Trig, Exponential — for which factor to call $u$. Higher class wins.

$dv = dx$ trickFor a single-function integrand like $\ln x$, take $u = \ln x$, $dv = dx$, so $v = x$.

Constant of integrationThe $+C$ added to indefinite integrals; in IBP it appears once, at the final answer.

MEX-C1NESA outcome (Further Integration): applies further integration techniques including integration by parts.

The integration by parts formula

core concept

Start from the product rule. If $u$ and $v$ are functions of $x$, then $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$. Integrate both sides with respect to $x$:

$$uv \;=\; \int u\,\frac{dv}{dx}\,dx \;+\; \int v\,\frac{du}{dx}\,dx \;=\; \int u\,dv \;+\; \int v\,du$$

Rearranging gives the working form used in every exam:

$$\int u\,dv \;=\; uv \;-\; \int v\,du$$

This replaces one integral with another. The technique only helps if the new integral $\int v\,du$ is easier than the original $\int u\,dv$. That's why choosing $u$ matters.

LIATE explained. Functions higher on the list (Log, Inverse trig) become much simpler when differentiated — $\ln x \to \tfrac{1}{x}$, $\tan^{-1} x \to \tfrac{1}{1+x^2}$. Functions lower on the list (Exponential, Trig) stay the same complexity when differentiated. So pick $u$ from the highest class present, leaving $dv$ to be something integrable.

IBP formula: $\int u\,dv = uv - \int v\,du$ · Derivation: integrate the product rule $(uv)' = u'v + uv'$ · LIATE order: L og, I nv trig, A lg, T rig, E xp — choose $u$ from the top · Always write a $u$/$dv$ table before substituting

Pause — copy the IBP formula $\int u\,dv = uv - \int v\,du$, its product-rule derivation, and the LIATE order (Log, Inv trig, Algebraic, Trig, Exp) into your book.

Quick check: For $\int x \cos x \, dx$, which choice of $u$ does LIATE recommend?

Choosing $u$ and $dv$ with LIATE

core concept

We just saw the IBP formula $\int u\,dv = uv - \int v\,du$, derived by integrating the product rule, with the LIATE hierarchy guiding the choice of $u$. That raises a question: how do we systematically apply LIATE to avoid choosing the wrong $u$? This card answers it → $u$ from the highest LIATE class present; $dv$ = everything else (including $dx$); verify $du$ simplifies and $v$ exists in closed form.

The same integrand can be split many ways — but only one choice produces a simpler integral. LIATE encodes the experience of generations of integrators.

Logarithmic: $\ln x$, $\log_a x$ — always pick as $u$ if present (and a polynomial is in $dv$).
Inverse trig: $\sin^{-1} x$, $\tan^{-1} x$ — pick as $u$.
Algebraic: $x$, $x^2$, polynomials — pick as $u$ over trig/exp.
Trig: $\sin x$, $\cos x$ — usually $dv$.
Exponential: $e^x$, $a^x$ — usually $dv$.

Quick sanity test. After choosing, ask: is $du$ simpler than $u$? Is $v = \int dv$ available? If both answers are yes, proceed. If $du$ is uglier or $\int dv$ won't close, swap.

$$\text{Standard IBP table:} \quad u = \square,\; du = \square\,dx; \quad dv = \square\,dx,\; v = \square$$

Common mistake. Forgetting that LIATE only prioritises $u$ — $dv$ is just "everything else, including $dx$". Students often try to write $dv = x$ without the $dx$. Always include the differential.

LIATE = Log, Inverse trig, Algebraic, Trig, Exponential · $u$ from highest class; $dv$ = everything else (with $dx$) · Check: $du$ simpler than $u$? $v$ available in closed form? · If product rule of the answer recovers the integrand, you're right

Pause — copy the LIATE rule in full, the two-step check ($du$ simpler? $v$ available?), and the verification method (differentiate $uv$ via product rule and confirm you recover the integrand) into your book.

Did you get this? True or false: for $\int x \ln x \, dx$, LIATE recommends $u = \ln x$ and $dv = x\,dx$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · $\int x \sin x \, dx$

Evaluate $\displaystyle \int x \sin x \, dx$ using integration by parts.

LIATE: Algebraic ($x$) beats Trig ($\sin x$), so $u = x$, $dv = \sin x\,dx$. Then $du = dx$ and $v = -\cos x$.

Differentiating $x$ gives $1$ (simpler); integrating $\sin x$ gives $-\cos x$ (no harder). Both checks pass.

PROBLEM 2 · $\int x e^x \, dx$

Evaluate $\displaystyle \int x e^x \, dx$.

LIATE: Algebraic beats Exponential, so $u = x$, $dv = e^x\,dx$. Then $du = dx$ and $v = e^x$.

$e^x$ is unique — its integral is itself. So $dv = e^x\,dx \Rightarrow v = e^x$ requires no work.

PROBLEM 3 · $\int \ln x \, dx$ (the $dv = dx$ trick)

Evaluate $\displaystyle \int \ln x \, dx$. There's no obvious product — invent one.

Write $\int \ln x\,dx = \int \ln x \cdot 1\,dx$. Set $u = \ln x$, $dv = dx$. Then $du = \dfrac{1}{x}\,dx$ and $v = x$.

LIATE: Log beats Algebraic, so $\ln x$ becomes $u$. With nothing else available, $dv = dx$ and $v = x$.

Fill the gap: The integration by parts formula is $\int u\,dv = $ $ - \int $ , derived by integrating the product rule.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Wrong choice of $u$ — making things worse

If you pick $u = \sin x$ in $\int x \sin x\,dx$, then $du = \cos x\,dx$ and $dv = x\,dx$ gives $v = \tfrac{x^2}{2}$. The new integral $\int \tfrac{x^2}{2}\cos x\,dx$ is harder than the original. LIATE prevents this by ranking Algebraic above Trig.

Trap 02

Dropping the minus sign in $-\int v\,du$

The formula is $uv - \int v\,du$, not $uv + \int v\,du$. Add a quick negative-sign check after substituting. This is the single most common arithmetic slip on IBP questions.

Trap 03

Forgetting $+C$

Indefinite integrals must end with $+C$. Markers deduct for missing constants of integration. Definite integrals don't need it (limits cancel it out), but indefinite ones always do.

Did you get this? True or false: $\displaystyle \int x e^x \, dx = x e^x + e^x + C$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Evaluate $\displaystyle \int x \cos x \, dx$ by parts. Use LIATE to choose $u$.

Evaluate $\displaystyle \int x e^{2x} \, dx$. Be careful with the $\tfrac{1}{2}$ that appears from integrating $e^{2x}$.

Evaluate $\displaystyle \int x \ln x \, dx$. (LIATE: Log beats Algebraic, so $u = \ln x$.)

Evaluate $\displaystyle \int_0^1 x e^{-x} \, dx$. Apply parts and then substitute the limits.

Evaluate $\displaystyle \int \tan^{-1} x \, dx$ using the $dv = dx$ trick.

Odd one out: Three of these integrals are routine first-application IBP problems. Which one requires repeated application of IBP?

Revisit your thinking

Earlier you started from the product rule and rearranged to obtain the IBP formula.

The formula $\int u\,dv = uv - \int v\,du$ doesn't evaluate integrals; it swaps one for another. The skill is choosing $u$ and $dv$ so the swap produces something easier. LIATE is the practical guide — Log and Inverse trig become much simpler when differentiated, so they're best as $u$. Algebraic factors should also become $u$ when paired with Trig or Exponential. The $dv = dx$ trick handles single-function integrands like $\ln x$. Once you've made the right split, the rest is bookkeeping.

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Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Evaluate $\displaystyle \int x \cos x \, dx$. Show your choice of $u$ and $dv$. (2 marks)

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ApplyBand 43 marks

Q2. Evaluate $\displaystyle \int_1^{e} \ln x \, dx$. State the IBP setup clearly. (3 marks)

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AnalyseBand 53 marks

Q3. Evaluate $\displaystyle \int x \ln x \, dx$. Justify your choice of $u$ using LIATE. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $u = x$, $dv = \cos x\,dx \Rightarrow du = dx$, $v = \sin x$. $\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C$.

2. $u = x$, $dv = e^{2x}\,dx \Rightarrow du = dx$, $v = \tfrac{1}{2}e^{2x}$. $\int x e^{2x}\,dx = \tfrac{1}{2}xe^{2x} - \tfrac{1}{2}\int e^{2x}\,dx = \tfrac{1}{2}xe^{2x} - \tfrac{1}{4}e^{2x} + C = \tfrac{1}{4}(2x - 1)e^{2x} + C$.

3. $u = \ln x$, $dv = x\,dx \Rightarrow du = \tfrac{1}{x}\,dx$, $v = \tfrac{x^2}{2}$. $\int x\ln x\,dx = \tfrac{x^2}{2}\ln x - \int \tfrac{x}{2}\,dx = \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C$.

4. $u = x$, $dv = e^{-x}\,dx \Rightarrow du = dx$, $v = -e^{-x}$. $\int_0^1 x e^{-x}\,dx = [-xe^{-x}]_0^1 + \int_0^1 e^{-x}\,dx = -e^{-1} + [-e^{-x}]_0^1 = -e^{-1} - e^{-1} + 1 = 1 - 2e^{-1}$.

5. $u = \tan^{-1} x$, $dv = dx \Rightarrow du = \tfrac{1}{1+x^2}\,dx$, $v = x$. $\int \tan^{-1} x\,dx = x\tan^{-1} x - \int \tfrac{x}{1+x^2}\,dx = x\tan^{-1} x - \tfrac{1}{2}\ln(1+x^2) + C$.

Q1 (2 marks): $u = x$, $dv = \cos x\,dx$ [1]; $\int x\cos x\,dx = x\sin x + \cos x + C$ [1].

Q2 (3 marks): $u = \ln x$, $dv = dx$ [1]; antiderivative $x\ln x - x$ [1]; $[x\ln x - x]_1^e = (e - e) - (0 - 1) = 1$ [1].

Q3 (3 marks): LIATE: Log beats Algebraic, so $u = \ln x$, $dv = x\,dx$ [1]; $\int x\ln x\,dx = \tfrac{x^2}{2}\ln x - \int \tfrac{x}{2}\,dx$ [1]; $= \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C$ [1].

Boss battle · The Parts Master

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Five timed questions on choosing $u$ and applying IBP. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering quick IBP questions. Lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 2