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 $\int \ln x\,dx$.
+5 XP warm-up Write out the product rule for differentiation: $\dfrac{d}{dx}(uv) = \;?$.
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?
+5 XP to read Every parts question rewards two habits: split the integrand into and (one to differentiate, one to...
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: L og · I nverse trig · A lgebraic · T rig · E xponential, pick $u$ from the highest class present.
Start from the product rule.
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 $x$:
Rearranging gives the working form used in every exam:
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.
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?
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 $u$.
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.
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.
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
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle \int x \sin x \, dx$ using integration by parts.
Evaluate $\displaystyle \int x e^x \, dx$.
Evaluate $\displaystyle \int \ln x \, dx$. There's no obvious product, invent one.
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
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\displaystyle \int x e^x \, dx = x e^x + e^x + C$.
Activities · practice with the ideas
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?
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.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $\displaystyle \int x \cos x \, dx$. Show your choice of $u$ and $dv$. (2 marks)
Q2. Evaluate $\displaystyle \int_1^{e} \ln x \, dx$. State the IBP setup clearly. (3 marks)
Q3. Evaluate $\displaystyle \int x \ln x \, dx$. Justify your choice of $u$ using LIATE. (3 marks)
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].
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
Work through this topic 1-on-1 with an experienced HSC tutor.
Book a free session →