IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~12 min read

Integration by Parts

Parts is the reverse-product-rule companion to reverse-chain-rule (substitution). When the integrand is a product of two functions and substitution doesn’t help, parts is the tool. The choice of u is the whole game — get it right and the integral simplifies, get it wrong and it gets worse. Use the LIATE mnemonic (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) to pick u: the function HIGHER on the list becomes u. Three variants to know: standard parts, repeated parts (for x²·something), and trap-and-solve (when the integral comes back to itself).

📘 What you need to know

Formula booklet: ∫ u dvdx dx = uv − ∫ v dudx dx, equivalently ∫ u dv = uv − ∫ v du.
LIATE for choosing u: L (ln), I (arcsin/arctan), A (algebraic), T (trig), E (exp). Function HIGHER on the list becomes u; the other becomes dv/dx.
Why LIATE works: u should become SIMPLER under differentiation. ln x simplifies to 1/x; polynomial xⁿ simplifies to nxⁿ⁻¹. Exp and trig don’t simplify under differentiation — they’re better as dv/dx.
Single functions via parts: ∫ln x dx, ∫arcsin x dx, ∫arctan x dx can ALL be done by parts. Rewrite as 1·f(x), then u = f(x), dv/dx = 1.
Repeated parts: if the new integral ∫v du is still a product (e.g., still polynomial × exp/trig), apply parts again. Common for ∫x² f(x) dx.
Trap and solve: for ∫e^kx·(sin or cos) dx, the original integral I reappears on the right after two applications. Treat I as an algebraic unknown and solve.
For v, no need for “+c“ at the intermediate step — any antiderivative works for the formula, and the +c only appears once, at the end of the final answer.
+c on every indefinite; for definite integrals, the [uv]_a^b term also gets evaluated at the limits.

The parts formula and LIATE

Integration by parts — formula booklet ∫ u dvdx dx = uv − ∫ v dudx dx or ∫ u dv = uv − ∫ v du

The formula is just the product rule rearranged — but the strategic challenge is choosing which factor is u. LIATE captures the intuition: pick u so that differentiating it makes the integrand simpler.

The integrand xe^x is a product of an algebraic factor and an exponential — neither is the derivative of the other, so substitution won’t help. LIATE picks u = x (A beats E), and one application of parts reduces ∫xe^x dx to ∫e^x dx, which is trivial. The antiderivative (x−1)e^x evaluates to 1 across [0, 1].

Why LIATE works: differentiating ln x gives 1/x (simpler); differentiating x³ gives 3x² (simpler); differentiating sin x gives cos x (same complexity); differentiating e^x gives e^x (no simpler). So we PREFER to differentiate the higher-LIATE function and integrate the lower-LIATE one. Make u the function you WANT to simplify.

LIATE priority and standard variants

LIATE — priority for choosing u

L — Logarithmic (ln x, log_a x)
I — Inverse trig (arcsin, arctan)
A — Algebraic (x, x², polynomials)
T — Trigonometric (sin, cos, tan)
E — Exponential (e^x, a^x)

Function HIGHER on the list becomes u; the other becomes dv/dx.

Three variants of parts

Standard — one application of parts
Repeated — apply parts twice for x² × f(x)
Trap & solve — for e^kx × (sin/cos), I reappears, solve algebraically

All use the same parts formula; the difference is how many times it’s applied and what happens at the end.

🧭 Recipe — integration by parts

Name the integral I and choose u, dv/dx: use LIATE — the function higher on the list becomes u, the other becomes dv/dx. For single functions like ln x, rewrite as 1·ln x and set u = ln x, dv/dx = 1.
Compute du/dx by differentiating u, and find v by integrating dv/dx. Take the simplest antiderivative for v — no need for +c at this stage.
Apply the formula I = uv − ∫v · du/dx dx. Watch the minus sign — it’s the most common source of errors.
Evaluate the new integral. If it’s straightforward, finish. If it’s still a product, apply parts AGAIN (repeated parts). If the original I reappears, treat I as an unknown and SOLVE algebraically (trap and solve).
Finish: for indefinite, add +c at the very end; for definite, evaluate [uv] at the limits AND apply limits to the remaining integral. Simplify the answer if possible — factoring out common terms often makes it cleaner.

Worked examples

WE 1

Standard parts — algebraic × exponential

Find ∫ x e^−2x dx.

Step 1 — name and choose u using LIATE I = ∫ x e^(-2x) dx A (x) vs E (e^(-2x)): A beats E, so u = x u = x, dv/dx = e^(-2x) Step 2 — differentiate u and integrate dv/dx du/dx = 1 v = ∫ e^(-2x) dx = -e^(-2x)/2 Step 3 — apply the parts formula I = uv – ∫ v · (du/dx) dx I = x · (-e^(-2x)/2) – ∫ (-e^(-2x)/2) · 1 dx I = -x e^(-2x)/2 + (1/2) ∫ e^(-2x) dx Step 4 — finish the new integral I = -x e^(-2x)/2 + (1/2)(-e^(-2x)/2) + c I = -x e^(-2x)/2 – e^(-2x)/4 + c Step 5 — factor for the cleanest form I = -e^(-2x)/4 · (2x + 1) + c ∫ x e^(-2x) dx = − (2x + 1) e^(-2x)4 + c verify: d/dx[-(2x+1)e^(-2x)/4] = -[2e^(-2x) – 2(2x+1)e^(-2x)]/4 = -e^(-2x)(2 – 4x – 2)/4 = x e^(-2x). ✓ classic example where A beats E in LIATE.

WE 2

Standard parts — algebraic × trigonometric

Find ∫ x sin(2x) dx.

Step 1 — choose u via LIATE A (x) vs T (sin 2x): A beats T, so u = x u = x, dv/dx = sin(2x) Step 2 — du/dx and v du/dx = 1 v = ∫ sin(2x) dx = -cos(2x)/2 Step 3 — apply the formula I = x · (-cos(2x)/2) – ∫ (-cos(2x)/2)(1) dx I = -x cos(2x)/2 + (1/2) ∫ cos(2x) dx Step 4 — finish the new integral I = -x cos(2x)/2 + (1/2) · sin(2x)/2 + c I = -x cos(2x)/2 + sin(2x)/4 + c Step 5 — common form I = (sin(2x) – 2x cos(2x))/4 + c ∫ x sin(2x) dx = sin(2x) − 2x cos(2x)4 + c v = -cos(2x)/2 has the 1/2 from “adjust and compensate” inside the integration of sin(2x). This 1/2 propagates through both terms — easy to drop if you’re not careful with the chain-rule constants.

WE 3

Single function — the 1·ln x trick

Find ∫ ln x dx.

Step 1 — single function trick no product visible — rewrite as ∫ 1 · ln(x) dx LIATE: L beats A — so u = ln x, dv/dx = 1 Step 2 — du/dx and v u = ln x ⟹ du/dx = 1/x dv/dx = 1 ⟹ v = x Step 3 — apply the formula I = uv – ∫ v · (du/dx) dx I = x ln(x) – ∫ x · (1/x) dx I = x ln(x) – ∫ 1 dx Step 4 — finish I = x ln(x) – x + c ∫ ln x dx = x ln x − x + c classic must-know result. Same trick gives ∫ arcsin x dx and ∫ arctan x dx — rewrite as 1·f(x), let u = f(x), dv/dx = 1. Works for ANY single function whose derivative is “nice” (like 1/x, 1/√(1-x²), 1/(1+x²)).

WE 4

Definite parts — clean integer answer

Evaluate ∫₀¹ x e^x dx.

Step 1 — choose u via LIATE A (x) vs E (e^x): A beats E, so u = x u = x, dv/dx = e^x Step 2 — du/dx and v du/dx = 1, v = e^x Step 3 — apply parts WITH limits I = [uv] from 0 to 1 – ∫ from 0 to 1 of v · (du/dx) dx I = [x e^x] from 0 to 1 – ∫ from 0 to 1 of e^x dx Step 4 — evaluate each piece [x e^x] from 0 to 1 = 1 · e – 0 · 1 = e ∫ from 0 to 1 of e^x dx = [e^x] from 0 to 1 = e – 1 Step 5 — combine I = e – (e – 1) = e – e + 1 = 1 ∫ from 0 to 1 of x e^x dx = 1 the e from the [xe^x] boundary cancels exactly the e from the second integral, leaving the clean integer 1. This is the integral plotted in the SVG above — the area under x·e^x from 0 to 1 is exactly 1 square unit.

WE 5

Repeated parts — x² × exponential

Find ∫ x² e^−x dx.

Step 1 — first parts, LIATE picks u = x² I = ∫ x² e^(-x) dx u = x², du/dx = 2x dv/dx = e^(-x), v = -e^(-x) Step 2 — apply the formula I = x² · (-e^(-x)) – ∫ (-e^(-x)) · 2x dx I = -x² e^(-x) + 2 ∫ x e^(-x) dx Step 3 — the new integral is STILL a product → apply parts AGAIN J = ∫ x e^(-x) dx u = x, du/dx = 1 dv/dx = e^(-x), v = -e^(-x) J = x · (-e^(-x)) – ∫ (-e^(-x))(1) dx J = -x e^(-x) + ∫ e^(-x) dx J = -x e^(-x) – e^(-x) + c (c will absorb) J = -(x + 1) e^(-x) + c Step 4 — substitute J back into I I = -x² e^(-x) + 2 · [-(x + 1) e^(-x)] + c I = -x² e^(-x) – 2(x + 1) e^(-x) + c Step 5 — factor I = -e^(-x) [x² + 2x + 2] + c ∫ x² e^(-x) dx = −(x² + 2x + 2) e^(-x) + c each round of parts brings the polynomial power down by 1: x² → 2x → 2 (constant). For ∫ x³ e^(-x) dx you’d need THREE applications. The “+c” only matters at the very end — drop it from intermediate J.

WE 6

Trap and solve — exp × trig

Find ∫ e^2x cos(x) dx.

Step 1 — name and apply parts. Neither factor simplifies; pick CONSISTENTLY I = ∫ e^(2x) cos(x) dx Choose u = e^(2x) (will keep this consistent on round 2) u = e^(2x), du/dx = 2 e^(2x) dv/dx = cos(x), v = sin(x) Step 2 — first application I = e^(2x) sin(x) – ∫ sin(x) · 2 e^(2x) dx I = e^(2x) sin(x) – 2 ∫ e^(2x) sin(x) dx Step 3 — second parts on J = ∫ e^(2x) sin(x) dx (same u choice) u = e^(2x), du/dx = 2 e^(2x) dv/dx = sin(x), v = -cos(x) J = e^(2x) · (-cos x) – ∫ (-cos x)(2 e^(2x)) dx J = -e^(2x) cos(x) + 2 ∫ e^(2x) cos(x) dx J = -e^(2x) cos(x) + 2I ← THE ORIGINAL I REAPPEARS Step 4 — substitute and solve algebraically I = e^(2x) sin(x) – 2J I = e^(2x) sin(x) – 2[-e^(2x) cos(x) + 2I] I = e^(2x) sin(x) + 2 e^(2x) cos(x) – 4I 5I = e^(2x) sin(x) + 2 e^(2x) cos(x) 5I = e^(2x) (sin(x) + 2 cos(x)) Step 5 — solve for I and add +c I = (1/5) e^(2x) (sin(x) + 2 cos(x)) + c ∫ e^(2x) cos(x) dx = e^(2x) (sin x + 2 cos x)5 + c consistency matters: same u choice (e^(2x)) on both rounds. Switching would give I = … – I/4 instead, which is wrong because the signs don’t trap the integral. Strict LIATE says u = cos x (T beats E), but for trap-and-solve either choice works — pick whichever gives cleaner algebra.

💡 Top tips

LIATE picks u in every case: Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential. The HIGHER function becomes u.
For single functions (ln x, arcsin x, arctan x), rewrite as 1·f(x) and let u = f(x), dv/dx = 1. Parts converts the integral into one involving f′(x), which is usually integrable.
For repeated parts, label your integrals (I, J) to avoid confusion. Each round reduces the polynomial degree by one — for x² you need TWO rounds, for x³ you need THREE.
For trap-and-solve, watch for I reappearing after two rounds. Use the SAME u choice in both rounds — switching causes I to cancel and gives 0 = (something), which is useless.
Verify by differentiating — for parts answers with multiple terms and constants, this is the only way to confirm signs and coefficients are right.

⚠ Common mistakes

Choosing u badly: setting u = e^x when a polynomial is present makes the integral harder, not easier. Trust LIATE.
Forgetting the minus sign in ∫u dv = uv − ∫v du. The minus is essential — writing + gives a fundamentally wrong answer.
Giving up too early on repeated parts: after one application, if the new integral is STILL a product (e.g., x² became x), apply parts AGAIN. Don’t conclude “this doesn’t work.”
Switching u mid-trap-and-solve: if you chose u = e^2x on round 1, you must do the same on round 2. Switching causes I to cancel and you’ll end up with 0 = expression.
Forgetting +c“ after the algebra of trap-and-solve. Solving “5I = stuff” gives I = stuff/5 — and then you MUST add +c at the very end.

Up next: Integrating with Partial Fractions. When the integrand is a rational function whose denominator factorises into linear factors, partial-fraction decomposition splits it into simpler pieces — each of which is just a 1/(ax+b) integral giving a ln antiderivative. The trick: first check if f′/f handles it (one line), THEN check if completing-the-square gives an inverse-trig form, and only THEN go to partial fractions. Three tools for rational integrands; partial fractions is the workhorse.

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