IB Maths AA HLTopic 5 — CalculusPaper 1 & 2HL~12 min read
Integrating with Partial Fractions
Partial fractions decompose a rational integrand P(x)/Q(x) into simpler pieces — each a fraction of the form A/(x − r) — so the integral becomes a sum of ln terms. The technique is powerful but should be your THIRD choice, not your first. Always check: (1) is the numerator (a multiple of) the derivative of the denominator? If yes, use the f′/f shortcut. (2) Does the denominator factorise? If no real roots, complete the square and use arctan. Only when both shortcuts fail and Q factors into linear pieces should you reach for partial fractions.
📘 What you need to know
What partial fractions does: writes P(x)(x − r)(x − s) as Ax − r + Bx − s. Each piece integrates to a ln antiderivative.
Decision tree (always check in order): (1) Is numerator = (multiple of) derivative of denominator? → use f′/f shortcut. (2) Does Q(x) factor? If NO real roots → complete the square, use arctan. (3) Otherwise → partial fractions.
Finding A and B — two methods: (i) “cover-up”: multiply both sides by (x − r), then substitute x = r to isolate A; (ii) simultaneous equations: equate coefficients of x and the constant on both sides.
Cover-up is faster for distinct linear factors. Set x = each root in turn — the other factor “covers up” and that constant pops out directly.
Integration of each piece: ∫ A/(x − r) dx = A ln|x − r| + c. Modulus essential unless f(x) provably positive on the interval.
Combining logs: A ln|f| + B ln|g| = ln|fA gB| — useful when a single-log final form is required. The constant of integration can be absorbed as ln(k) if you want a really compact form.
For definite integrals, you can keep the partial-fraction sum form or combine first — both work. Apply the upper-minus-lower limit evaluation as usual.
Discriminant tells you the route: b² − 4ac > 0 → factors → partial fractions; < 0 → no real roots → complete the square + arctan; = 0 → repeated root → out of syllabus territory for AA HL.
The decision tree
Three routes lead to a clean answer for a rational integrand. Before computing anything, check which route applies — it can save many minutes of unnecessary algebra.
Three routes for any rational integrand P(x)/Q(x) with quadratic Q. Always check the f′/f shortcut FIRST (one line). Then check the discriminant of Q. Reach for partial fractions only when neither shortcut applies AND Q factors into linear pieces — that’s the third-priority route, not the default.
Three integrals, three answers — same integrand shape: ∫ (2x+3)/(x²+3x+5) dx uses f′/f → ln(x²+3x+5). ∫ 1/(x²+2x+5) dx needs complete-the-square → (1/2)arctan((x+1)/2). ∫ 1/(x²−4) dx needs partial fractions → (1/4)ln|(x−2)/(x+2)|. The integrand SHAPE looks similar but the antiderivative TYPE depends entirely on the discriminant and whether numerator matches Q′.
Cover-up vs simultaneous equations
Cover-up method (faster)
P(x)(x − r)(x − s) = Ax−r + Bx−s
A = P(r)/(r−s), B = P(s)/(s−r)
Substitute x = each root in turn into the numerator-equals-numerator identity. The other factor vanishes (“covers up”), isolating A or B directly.
Simultaneous equations (more general)
Multiply out: P(x) = A(x−s) + B(x−r)
Equate coefficients of x and constant on both sides → solve 2×2 system for A, B
Works for any decomposition; required when the question explicitly asks for it, or when one method gives messy fractions and the other is cleaner.
🧭 Recipe — integrate any P(x)/Q(x) with Q quadratic
Check f′/f FIRST: is the numerator (a constant multiple of) the derivative of the denominator? If yes, the integral is (constant)·ln|denominator| + c in one line. Skip partial fractions entirely.
Check the discriminant of Q(x): if b² − 4ac < 0, the denominator has NO real roots. Complete the square to get (x+k)² + a² and use the arctan formula — NOT partial fractions.
If Q factors (discriminant > 0), write P(x)/Q(x) = A/(x−r) + B/(x−s) where r, s are the roots of Q. Find A and B via cover-up or simultaneous equations.
Integrate each piece: ∫ A/(x−r) dx = A ln|x−r| + c. Sum the pieces to get the full antiderivative.
Finish: for indefinite, add +c and (if asked for single-log form) combine using log laws. For definite, apply upper-minus-lower at the limits — log laws often simplify the difference into a single ln.
Worked examples
WE 1
Basic partial fractions — difference of squares
Find ∫ 1x² − 4 dx.
Step 1 — decision checknumerator = 1; derivative of (x²-4) is 2x — NOT a match for f’/fdiscriminant of x²-4: 0² – 4(1)(-4) = 16 > 0 — Q factors → partial fractionsStep 2 — factor and decomposex² – 4 = (x-2)(x+2)1/((x-2)(x+2)) = A/(x-2) + B/(x+2)Step 3 — cover-up methodmultiply both sides by (x-2)(x+2):1 = A(x+2) + B(x-2)x = 2: 1 = 4A ⟹ A = 1/4x = -2: 1 = -4B ⟹ B = -1/4Step 4 — integrate each pieceI = ∫ (1/4)/(x-2) dx – ∫ (1/4)/(x+2) dx = (1/4) ln|x-2| – (1/4) ln|x+2| + cStep 5 — combine logs = (1/4) ln|(x-2)/(x+2)| + c∫ 1/(x² − 4) dx = 14 ln|x − 2x + 2| + cclassic “difference of squares” partial fractions — uses the identity 1/(a²-b²) = (1/(2a)) · [1/(a-b) – 1/(a+b)] · … well, more directly, just cover-up with the two roots.
WE 2
Non-constant numerator — full decomposition
Find ∫ 2x + 5x² − x − 2 dx.
Step 1 — decision checkderivative of (x²-x-2) is 2x-1; numerator is 2x+5 ≠ multiple of 2x-1 → not f’/fdiscriminant: 1 + 8 = 9 > 0 → factors → partial fractionsStep 2 — factor and decomposex² – x – 2 = (x-2)(x+1)(2x+5)/((x-2)(x+1)) = A/(x-2) + B/(x+1)Step 3 — cover-up2x + 5 = A(x+1) + B(x-2)x = 2: 9 = 3A ⟹ A = 3x = -1: 3 = -3B ⟹ B = -1Step 4 — integrateI = ∫ 3/(x-2) dx + ∫ -1/(x+1) dx = 3 ln|x-2| – ln|x+1| + cStep 5 — combine into single log = ln|(x-2)³| – ln|x+1| + c = ln|(x-2)³/(x+1)| + c∫ (2x+5)/(x²−x−2) dx = ln|(x−2)³x+1| + ccover-up handles a non-constant numerator just as easily — substitute each root, the (other factor) “covers up” and the constant pops out. The “3 ln|x-2|” combines using the log law A ln|f| = ln|f^A| into ln|(x-2)³|.
WE 3
Definite integral via partial fractions
Evaluate ∫341x² − 1 dx, giving your answer in exact form.
Step 1 — decision check; factorx² – 1 = (x-1)(x+1)numerator 1 ≠ Q'(x) = 2x; → partial fractionsStep 2 — decompose by cover-up1 = A(x+1) + B(x-1)x = 1: 1 = 2A ⟹ A = 1/2x = -1: 1 = -2B ⟹ B = -1/2Step 3 — antiderivative (no need for +c on a definite)F(x) = (1/2) ln|x-1| – (1/2) ln|x+1| = (1/2) ln|(x-1)/(x+1)|Step 4 — apply limits 3 to 4F(4) = (1/2) ln|3/5|F(3) = (1/2) ln|2/4| = (1/2) ln(1/2)Step 5 — compute the difference using log lawF(4) – F(3) = (1/2)[ln(3/5) – ln(1/2)] = (1/2) ln[(3/5)/(1/2)] = (1/2) ln(6/5)∫ from 3 to 4 of 1/(x²−1) dx = 12 ln(65)interval [3, 4] is well inside x > 1, so x-1 and x+1 are both positive — modulus drops naturally. Final answer is positive (since 6/5 > 1, so ln(6/5) > 0) — matches the fact that 1/(x²-1) > 0 on [3, 4].
WE 4
Decision tree — spot f′/f, DON’T use partial fractions
Find ∫ 2x + 3x² + 3x + 5 dx.
Step 1 — the decision-tree checkdenominator: f(x) = x² + 3x + 5f'(x) = 2x + 3 ← exactly the numerator!⟹ this is f'(x)/f(x) — use the SHORTCUT, NOT partial fractionsStep 2 — apply f’/f formula∫ f'(x)/f(x) dx = ln|f(x)| + cI = ln|x² + 3x + 5| + cStep 3 — can we drop the modulus?check discriminant: 9 – 20 = -11 < 0no real roots — f(x) doesn’t change sign anywheref(0) = 5 > 0 ⟹ f(x) > 0 for all x — drop modulus∫ (2x+3)/(x²+3x+5) dx = ln(x² + 3x + 5) + cEVEN THOUGH this looks like a partial-fractions setup (rational integrand with quadratic denominator), the f’/f match saves us all the decomposition work. Always do the f'(x) check first — it’s just one differentiation, and it often eliminates the harder algebra entirely.
WE 5
Decision tree — no real roots, use complete-the-square + arctan
Find ∫ 1x² + 2x + 5 dx.
Step 1 — decision tree checknumerator 1; derivative of x²+2x+5 is 2x+2 — not f’/fdiscriminant: 4 – 20 = -16 < 0 — NO real roots → NOT partial fractions⟹ complete the square and use arctanStep 2 — complete the squarex² + 2x + 5 = (x² + 2x + 1) + 4 = (x + 1)² + 4 = (x + 1)² + 2²Step 3 — apply ∫ 1/(a² + u²) du = (1/a) arctan(u/a), with u = x+1, a = 2I = ∫ 1/((x+1)² + 4) dx = (1/2) arctan((x+1)/2) + c∫ 1/(x²+2x+5) dx = 12 arctan(x + 12) + ctrap to avoid: trying partial fractions on x²+2x+5 leads to dead ends because the quadratic doesn’t factor over the reals (you’d get complex roots). The discriminant check at step 1 saves you from wasted work — always do it BEFORE trying to factor.
WE 6
Full decomposition with log-law combining
Find ∫ 3x − 4x² − 3x + 2 dx.
Step 1 — decision checkdenominator’s derivative: 2x – 3numerator 3x – 4: NOT a multiple of 2x – 3⟹ not f’/f. Check factorisation:x² – 3x + 2 = (x-1)(x-2). roots are 1 and 2 ⟹ partial fractionsStep 2 — decompose(3x-4)/((x-1)(x-2)) = A/(x-1) + B/(x-2)3x – 4 = A(x-2) + B(x-1)Step 3 — cover-upx = 1: -1 = -A ⟹ A = 1x = 2: 2 = B ⟹ B = 2Step 4 — integrate each pieceI = ∫ 1/(x-1) dx + ∫ 2/(x-2) dx = ln|x-1| + 2 ln|x-2| + cStep 5 — combine using log laws into single ln = ln|x-1| + ln|(x-2)²| + c = ln|(x-1)(x-2)²| + c∫ (3x−4)/(x²−3x+2) dx = ln|(x−1)(x−2)²| + cboth forms ln|x-1| + 2 ln|x-2| + c and ln|(x-1)(x-2)²| + c are fully correct — combine only if “as a single log” or “in simplified form” is asked. The coefficient 2 in front of ln|x-2| becomes the power on (x-2) when combining.
💡 Top tips
Always check the decision tree FIRST: f′/f shortcut → partial fractions → complete the square. Doing the discriminant + f′ check at the start saves minutes of pointless algebra.
Cover-up is faster than simultaneous equations for distinct linear factors. Substitute each root and the other factor vanishes — A and B pop out directly.
Keep the modulus on logs unless you can prove the argument is strictly positive on the integration interval. For definite integrals where the interval is well inside one sign region, the modulus drops naturally.
Combine logs with care: A ln|f| + B ln|g| = ln|fA gB|. Only combine if the question asks for a single log — otherwise the sum form is fine.
Discriminant ≥ 0 means partial fractions are POSSIBLE; discriminant < 0 means you MUST complete the square. Check the discriminant before reaching for partial-fraction algebra.
⚠ Common mistakes
Reaching for partial fractions when f′/f works: ∫(2x+3)/(x²+3x+5) dx is ONE line by f′/f. Doing partial fractions on a non-factoring quadratic is impossible anyway, but doing it on a factoring one wastes time.
Trying partial fractions when no real roots exist: x²+2x+5 has discriminant −16, so it doesn’t factor over the reals. Force-decomposing leads to complex coefficients (off-syllabus).
Forgetting the modulus on ln antiderivatives: ∫1/(x−2) dx = ln|x−2| + c, NOT ln(x−2) + c. Only drop the modulus when you can prove positivity.
Sign errors in cover-up: substituting x = −1 into A(x−2) gives A(−3) = −3A, not +3A. Careful arithmetic on the signs.
Forgetting the +c on indefinite integrals — even when log-law combining produces a single ln, the +c still belongs there.
Up next: Area Between Curve & y-axis. So far you’ve integrated to find areas bounded by a curve and the x-axis between two x-limits. This sub-chapter flips the orientation: when the area is bounded by the curve, the y-axis, and two y-limits (horizontal lines), you integrate with respect to y instead. Rearrange y = f(x) into x = g(y) and apply ∫|x| dy. Beware: areas to the LEFT of the y-axis are “negative” and need absolute-value treatment, just like with the x-axis version.
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