Splitting an Algebraic Fraction into Separate Terms Before Integrating

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Why a fraction under an integral sign needs an extra step first

Spec 5.2 gives a rule for one shape of expression: "Integration of $x^n$ and related sums, differences and constant multiples (excluding $n = -1$ and related sums/differences/multiples)." Read that precisely. The rule covers $x^n$ , and it covers anything built by adding, subtracting or scaling terms that are already individually shaped like $x^n$ . It says nothing at all about a fraction, because a fraction — two expressions divided by each other — is not automatically shaped like $x^n$ , or like a sum of things shaped like $x^n$ . $\frac{4x^3-6}{x^2}$ looks like one object on the page, but until it is rewritten it is not the kind of object spec 5.2's rule applies to.

The rewrite itself uses nothing beyond arithmetic you already trust. A fraction with several terms on top and one term on the bottom splits apart exactly the way it would recombine: $\frac{A+B}{D} = \frac{A}{D} + \frac{B}{D}$ , because adding two fractions that already share a denominator is just adding their numerators — this is that step run backwards. Once split, each piece is a single power of x divided by another single power of x, and the index law $x^m \div x^n = x^{m-n}$ (spec 1.1) turns each piece into a pure power of x on its own. Only then does spec 5.2's rule have something to attach to — term by term, not to the fraction as a whole.

A real Jan 2023 examiner report on exactly this question type names the first way this goes wrong before any integration even starts: "some candidates had no idea how to write the expression as the sum of two terms." Not a wrong split — no split at all. The fraction stayed a fraction, and nothing in spec 5.2 covers that shape.

One scope check is worth building into the habit from the start, because it comes straight from the spec's own wording, not from an examiner report: spec 5.2 explicitly EXCLUDES $n = -1$ — $\int x^{-1}\,dx$ needs , which is not P1 content. After splitting a fraction, check that none of the resulting terms is $x^{-1}$ (for instance, a numerator's constant term divided by a denominator of $x^1$ produces exactly that excluded case). A genuine WMA11 question will never be built so that a correct split lands on $x^{-1}$ — if your own split does, the split itself is the thing to re-check, not the integration rule.

Mechanism

"Integrate the top, integrate the bottom, then divide" is not risky — it is provably a different calculation

This is not a case of one method being safer than another; the two processes genuinely compute different things, and a proves it without needing any exam-specific evidence at all. Take the simplest possible case: $\int \frac{4x}{x}\,dx$ . Split first, the correct way: $\frac{4x}{x} = 4$ (a constant, since x cancels), so $\int 4\,dx = 4x + c$ — a function that genuinely depends on x, which it must, since the thing being integrated does not vanish. Now the wrong way: integrate the numerator and denominator as if they were separate, unrelated expressions, then divide the two results. $\int 4x\,dx = 2x^2$ ; $\int x\,dx = \frac{x^2}{2}$ ; dividing gives $\frac{2x^2}{x^2/2} = 4$ — a bare constant, with the $x^2$ cancelling out of the ratio entirely. That result cannot be right on its face: it claims $\frac{dy}{dx}$ was a flat, unchanging number when the original expression, $\frac{4x}{x}$ , so obviously simplifies to the constant 4 that even a correct calculation should have produced exactly that — and by coincidence of this simple example it does, which is exactly what makes it a fair test: even when the two methods happen to agree on a trivial case, the wrong method reaches its answer by an argument that would give a nonsense, x-dependent-looking ratio on almost any other fraction. Try it on $\int \frac{x^2}{x}\,dx$ instead — correctly, $\frac{x^2}{x}=x$ , so $\int x\,dx = \frac{x^2}{2}+c$ ; the wrong way, $\int x^2\,dx \div \int x\,dx = \frac{x^3}{3} \div \frac{x^2}{2} = \frac{2x}{3}$ — a completely different function of x, not even close to $\frac{x^2}{2}$ . This is the real content behind a real Jan 2023 examiner report's finding that this approach "resulted in no marks": it is not a partial-credit method that happens to be marked harshly, it is a calculation that does not compute the integral of the original expression at all.

Worked, in full

$\frac{dy}{dx} = \frac{6x^3 - 8}{2x^2}$ , curve through $(2, 10)$ : find y — splitting a two-term numerator before any integration happens (VERIDIAN-original numbers, built to match the real quoted shape — a fraction that splits into "the sum of two terms" — a phrase this lesson's own header quotes verbatim from the Jan 2023 Q3 examiner report; not a reproduction of that question, whose original fraction this research pass did not recover)

01
Before integrating anything, rewrite the fraction as a sum of separate powers of x. $\frac{dy}{dx}$ is not $x^n$ for a single n, and spec 5.2's rule only applies once it is: split term by term, $\frac{6x^3 - 8}{2x^2} = \frac{6x^3}{2x^2} - \frac{8}{2x^2}$ .
Earns: No mark of its own yet — this is exactly the step a real Jan 2023 examiner report records some candidates never taking at all: "some candidates had no idea how to write the expression as the sum of two terms." Every mark below depends on reaching this line first.
02
Apply $x^m/x^n = x^{m-n}$ to each piece, dividing the coefficients separately from the powers: $\frac{6x^3}{2x^2} = 3x^{3-2} = 3x$ ; $\frac{8}{2x^2} = 4x^{0-2} = 4x^{-2}$ . So $\frac{dy}{dx} = 3x - 4x^{-2}$ — two pure powers of x, ready for spec 5.2's integration rule.
Earns: M1 A1 — M1 for attempting to rewrite the fraction as a sum of terms with the index law applied to at least one piece (the real Jan 2023 Q3 mark scheme's own wording: "one processed index correct" is enough for this mark, so a wrong-but-consistent split still earns it, provided the working from here on is otherwise sound). A1 for the fully correct, simplified split shown here — a separate, additional mark under the real scheme, not folded into the M1.
03
Integrate each term separately, checking the excluded case first: spec 5.2 explicitly excludes $n = -1$ , and neither term here is $x^{-1}$ ( $x^1$ and $x^{-2}$ are both safe), so the ordinary rule applies to both. $\int 3x\,dx = \frac{3}{2}x^2$ ; $\int -4x^{-2}\,dx = -4 \cdot \frac{x^{-1}}{-1} = 4x^{-1}$ .
Earns: dM1 — the method mark for integration, earned the moment the power of at least one term increases by 1 ( $x^n \to x^{n+1}$ ), DEPENDENT on the split's M1 above (not on its A1) — the real Jan 2023 Q3 mark scheme states this explicitly for the identical step: "Depends on the first M mark." A wrong-but-consistent split (M1 earned, A1 not) can still unlock this mark; a split abandoned altogether cannot.
04
Collect the two integrated terms and add ONE constant of integration — one for the whole indefinite integral, not one per term: $y = \frac{3}{2}x^2 + 4x^{-1} + c$ .
Earns: No mark of its own yet — writing the symbolic $+c$ here is what keeps a free constant available for the point below to determine. Omitting c at this exact line — not later, right here — is the single most repeated "silly" mark loss on integration questions across the whole WMA11 corpus researched for this course, independently confirmed in four separate series (Jan 2023 Q3, Oct 2021 Q1, Oct 2020 Q9, Jan 2025 Q6(b)): skip this line, and the next one has nothing left to solve for.
05
Use the given point $(2, 10)$ to find c: $y(2) = \frac{3}{2}(4) + 4(2)^{-1} + c = 6 + 2 + c = 8 + c$ , and this equals 10, so $c = 2$ . $y = \frac{3}{2}x^2 + 4x^{-1} + 2$ , or equivalently $y = \frac{3}{2}x^2 + \frac{4}{x} + 2$ .
Earns: dM1 A1 — dM1 for substituting the given point and solving for c, DEPENDENT on the integration mark above (the real Jan 2023 Q11(b) mark scheme's own wording, for the same "integrate then apply a point" step: "Applies f(4) = 3 and solves to find a constant of integration. Depends on first M1"). A1 for the finished equation with c correctly calculated — ONE combined accuracy mark for the whole finished answer, not a separate mark for the value of c and another for the final expression; the real mark scheme has no line that splits those two. Check by substitution: at $x=2$ , $\frac{3}{2}(4) + \frac{4}{2} + 2 = 6 + 2 + 2 = 10$ ✓.

Source — Examiner report, Jan 2023

"A small minority attempted to integrate all 3 terms separately before division which resulted in no marks"

In your own words

In one sentence: why does "integrate the numerator, integrate the denominator, then divide" give a genuinely different — and wrong — answer from integrating the correctly split expression, rather than just a slower route to the same one?

Complete it yourself

Complete the chain — $\frac{dy}{dx} = \frac{3x^4 - 5}{x^2}$ , curve through $(1, 10)$ : find y

01
This is a fraction, not a single power of x — spec 5.2's integration rule cannot be applied to it directly yet. Split it into a sum of two terms first: $\frac{3x^4}{x^2} - \frac{5}{x^2}$ .
Earns: No mark of its own yet — every mark below depends on reaching a valid split first, the same dependency the worked-chain above makes explicit.
02
Apply $x^m/x^n = x^{m-n}$ to each piece separately: $\frac{3x^4}{x^2} = 3x^2$ ; $\frac{5}{x^2} = 5x^{-2}$ . So $\frac{dy}{dx} = 3x^2 - 5x^{-2}$ — two pure powers of x, ready to integrate termwise.
Earns: M1 A1 — M1 for attempting the rewrite with the index law applied to at least one term (the real Jan 2023 Q3 mark scheme's own wording: "one processed index correct" suffices); A1 for the fully correct, simplified split shown here.

Marked, line by line

$\frac{dy}{dx} = \frac{8x^5 - 6x^2}{2x^2}$ . Given that the curve $y = f(x)$ passes through the point $(1, 3)$ , find y in terms of x. (5) — VERIDIAN-original question; not a reproduction of any past-paper question, though the real Jan 2023 Q3's own fraction and full mark scheme are now known (see this file's header) and were used to check the mark-code structure below. The 5-mark tariff mirrors the real Jan 2023 Q3's own verified tariff (§2's per-question mark list for that series, "5,5,5,4,6,10,10,8,4,10,8" — position 3 is 5). The per-line mark allocations below are modelled on verified mark-scheme conventions — the split earns its own M1 (attempt) and A1 (accuracy) as two separate marks, the integration mark is dM1 (dependent on the split's M1, not independent), and applying the given point is a further dM1/A1 pair, cross-checked against the real Jan 2023 Q3 and Q11(b) mark schemes — not copied from a real scheme, which for an original question does not exist.

5 marks available

01
$\frac{8x^5 - 6x^2}{2x^2} = \frac{8x^5}{2x^2} - \frac{6x^2}{2x^2} = 4x^3 - 3$
M1 for attempting to rewrite the fraction as a sum of terms with the index law applied to at least one piece — the real Jan 2023 Q3 mark scheme's own wording: "one processed index correct" is enough for this mark. A1 for the fully correct, simplified split shown here — a separate, additional mark under the real scheme, not folded into the M1. Note the second term reduces to a plain constant (x⁰), not a negative power — splitting a fraction does not always produce negative indices, only when the denominator's power exceeds the numerator term's own.
M1 A1
02
$\int (4x^3 - 3)\,dx = x^4 - 3x + c$
Method mark for integration: power of at least one term increased by 1 (x³ → x⁴, and x⁰ → x¹ for the −3x term), DEPENDENT on the split's M1 above, not on its A1 — the real Jan 2023 Q3 mark scheme states this explicitly for the identical step: "Depends on the first M mark." Earned regardless of what the constant of integration turns out to be, or even whether one is written yet.
dM1
03
$y(1) = 1 - 3 + c = -2 + c = 3 \Rightarrow c = 5$ , so $y = x^4 - 3x + 5$
dM1 for substituting the given point and solving for c, DEPENDENT on the integration mark above — the real Jan 2023 Q11(b) mark scheme's own wording for this exact step, integrate-then-apply-a-point: "Applies f(4) = 3 and solves to find a constant of integration. Depends on first M1." A1 for the finished equation with c correctly calculated — ONE combined accuracy mark for the whole finished answer; the real mark scheme has no separate line for "the value of c" and another for "the final expression." Check: differentiating gives $4x^3-3$ , and $\frac{8x^5-6x^2}{2x^2}=4x^3-3$ ✓; at $x=1$ , $1-3+5=3$ ✓.
dM1 A1

Named traps

no-idea-how-to-split-the-fraction: Confirmed directly on the cleanest fully-worked example this research pass found: "some candidates had no idea how to write the expression as the sum of two terms" (Jan 2023, Q3). The fraction is left as one object because nothing in spec 5.2's rule looks like it applies to a fraction — the fix is mechanical, not conceptual: $\frac{A+B}{D} = \frac{A}{D} + \frac{B}{D}$ is fraction addition run backwards, and each resulting piece reduces to a pure power of x via $x^m/x^n = x^{m-n}$ (spec 1.1). Practise the split as its own separate step, before integration ever enters the working.
numerator-and-denominator-integrated-before-dividing: Confirmed on the same question: "A small minority attempted to integrate all 3 terms separately before division which resulted in no marks" (Jan 2023, Q3). This is not a wrong split that still earns partial credit — it is not a split at all, and it computes something genuinely different from the fraction's actual integral (the mechanism block above proves this on a simple case, not just describes it). Unlike an arithmetic slip inside a correct method, there is no method mark here to fall back on.
constant-of-integration-omitted: The single most common "silly" final-mark loss on integration questions across the whole WMA11 corpus reviewed for this course — independently confirmed as a repeated finding across FOUR separate series: Jan 2023 Q3, Oct 2021 Q1, Oct 2020 Q9, and Jan 2025 Q6(b). Spec 5.1 requires a constant of integration every time indefinite integration is performed, whether or not the question ever gives enough information to find its numeric value. Write $+c$ the instant the integration is finished, not after the point has been substituted — an expression with no free constant left in it has nowhere for a given point to attach.
wrong-split-can-still-earn-the-integration-method-mark: A real, quoted fragment from the Jan 2023 Q3 examiner report records three different wrong ways students split the actual question's fraction: "8x³+6x⁻² or 4x⁵+6x⁻² or 2x⁵+(3/2)x⁻². These candidates could still gain the second method mark for appropriate integration." The real mark scheme explains exactly why: the integration mark is coded dM1 — "Depends on the first M mark" (the split's own M1, which only needs "one processed index correct" to be earned), not on the split's separate A1 (full accuracy). A wrong-but-genuine split attempt still earns the split's M1, and that alone is enough to unlock the dependent integration mark — a wrong division, honestly committed to and then integrated correctly, is not the same failure as trap two above. The practical lesson: if you are unsure your split is right, do not panic, guess randomly, or leave the question blank — commit to a split and integrate it correctly, because the dependent mark only checks that a genuine method attempt came first, not that it was error-free.
correct-integration-left-unsimplified-or-missing-the-x: A second, separate way to lose the final accuracy mark even after splitting AND integrating correctly — newly confirmed by this lesson's own primary-source re-verification, not previously recorded in this course's facts bank: the real Jan 2023 Q3 examiner report continues past the +c trap to name it directly: "marks were lost unnecessarily for example by... not simplifying fully... [or by] other careless mistakes such as missing out 'x' but writing the index... which also cost the final accuracy mark." Concretely, writing the correct power as a bare exponent with no base — e.g. "⁻¹" floating on its own instead of $x^{-1}$ attached to its coefficient — states a number, not the term the mark scheme actually wants. The fix is procedural, not conceptual: once a term is integrated, check that the base variable and its exponent are both actually written down together, and that the whole expression is collected into its simplest single form, before moving on to substitute a point.

Retrieval — with feedback on every choice

Question 1

3 marks

Write $\frac{5x^4 + 2}{x^3}$ as the sum of two separate terms, each a power of x.

Question 2

1 mark

$\frac{dy}{dx} = \frac{x^3-2}{x^2}$ . Once split, why can this be integrated term-by-term using spec 5.2's rule, with no technique beyond P1 needed?

Question 3

2 marks

Which of these, once split into separate powers of x, produces a term that needs a technique beyond what P1 covers?

Question 4

2 marks

$\frac{dy}{dx} = \frac{4x^3-9}{x^2}$ and the curve $y=f(x)$ passes through $(1,2)$ . A student splits and integrates correctly to reach $y = 2x^2 + 9x^{-1} + c$ , then substitutes the point to find $c=-9$ , giving the final answer $y = 2x^2 + 9x^{-1} - 9$ . What did this student do right that the research bank's own four-series evidence says many real candidates get wrong?

Reference — not a study method, a lookup

Fraction under an integral sign? SPLIT first: (A+B)/D = A/D + B/D, then xᵐ/xⁿ = xᵐ⁻ⁿ on each piece. There is no rule for integrating a fraction as one object.
Never integrate numerator and denominator separately then divide — that computes a different quantity, not a slower route to the right one, and real mark schemes give it zero.
A wrong split can still earn the integration method mark, if what follows is integrated correctly and consistently — don't abandon a question over an uncertain split.
Every indefinite integration needs its OWN +c, written the moment you integrate — not deferred until a point is used. Confirmed the single most repeated silly mark loss across FOUR series (Jan 2023 Q3, Oct 2021 Q1, Oct 2020 Q9, Jan 2025 Q6(b)).
∫xⁿ dx = xⁿ⁺¹/(n+1) + c, n ≠ −1 — check no split term is x⁻¹; that case needs ln, outside P1, and real questions avoid it.

Not affiliated with or endorsed by Pearson Edexcel. Every fraction, point and numeric answer in this lesson's worked-chain, chain-drill and marked-solution is VERIDIAN-original — built to match the real, quoted SHAPE the Jan 2023 Q3 examiner report describes (a multi-term numerator over a single monomial denominator, splitting into "the sum of two terms," and the question's own real 5-mark tariff, taken from §2's verified per-question mark list for that series) but not a reproduction of Q3 itself: a mark-scheme-bullet coverage audit on 2026-09-14 re-extracted Q3's own real fraction, ∫(4x⁵+3)/(2x²)dx, and its full verbatim mark scheme directly from the primary Pearson PDFs (WMA11_01_MS_2301), and the worked-chain/chain-drill/marked-solution below still don't reproduce it, because Q3 has no point-substitution step and this lesson deliberately tests spec 5.1 (finding c from a point) alongside spec 5.2 (splitting) in one example. That same audit DID bring the M/A/B mark-code structure below into line with the real, now fully-known Q3 and Q11(b) mark schemes (M1/A1 for the split as two separate marks, dM1 — not independent M1 — for the integration that depends on it, dM1/A1 for applying the point). Three real quotations anchor this lesson's teach, mechanism and trap-taxonomy content, all from the same Jan 2023 Q3 examiner report, all checked directly against the primary Pearson document. This topic's own evidence is genuinely thinner than the two integration-adjacent topics the original six WMA11 lessons could draw on: the splitting-specific traps rest on that one fully-quoted series, not two or three — only the constant-of-integration trap is independently confirmed across four series (Jan 2023 Q3, Oct 2021 Q1, Oct 2020 Q9, Jan 2025 Q6(b)), which is why it alone gets a dedicated trap-taxonomy item. Per-line mark allocations on every original question are modelled on verified mark-scheme conventions (what M, dM and A marks mean and depend on, cross-checked against Q3's own split-then-integrate dependency chain and Q11(b)'s own integrate-then-apply-a-point dependency chain), not transcribed from a real scheme, which for an original question does not exist.

Question 13 marks

Write $\frac{5x^4 + 2}{x^3}$ as the sum of two separate terms, each a power of x.

$5x + 2x^{-3}$
Correct. $\frac{5x^4}{x^3} = 5x^{4-3} = 5x$ , and $\frac{2}{x^3} = 2x^{0-3} = 2x^{-3}$ — and $-3$ is not the excluded case $n=-1$ , so both terms integrate with the ordinary spec 5.2 rule.
B $5x^4 + 2x^{-3}$
The second term is right and the first is untouched. $\frac{5x^4}{x^3}$ still has to have the index law $x^m/x^n=x^{m-n}$ applied — leaving the first term as $5x^4$ is the same as never dividing it by the denominator at all.
C $5x - 2x^{-3}$
Both exponents are right and a sign has flipped. The original numerator is $5x^4 + 2$ , so the split preserves the $+$ : $5x + 2x^{-3}$ , not $5x - 2x^{-3}$ .
D $5x + 2x^{-2}$
The first term is right and the second exponent is wrong. $\frac{2}{x^3}$ needs the full power of the denominator subtracted, $0-3=-3$ , not $-2$ — this only accounts for part of the division.

Traps tested: Index law not applied to one term · Sign lost in the split · Index law subtraction error

Question 21 mark

$\frac{dy}{dx} = \frac{x^3-2}{x^2}$ . Once split, why can this be integrated term-by-term using spec 5.2's rule, with no technique beyond P1 needed?

Because the split gives $x - 2x^{-2}$ , and neither exponent is $-1$ — spec 5.2's rule excludes only $n=-1$ , and both these terms fall outside that exclusion.
Correct. The spec states the exclusion precisely: "excluding n = −1 and related sums/differences/multiples." The rule is checked term by term, and here it clears both terms.
BBecause both terms in the split are positive powers of x.
One of the two terms, $-2x^{-2}$ , is a negative power — and it still integrates fine, because the excluded case is specifically $n=-1$ , not negative powers in general. $\int x^{-2}\,dx = \frac{x^{-1}}{-1}+c$ is ordinary P1 content.
CBecause the fraction has already been simplified to remove the denominator entirely.
The denominator has not been removed — it has been absorbed into each term's own exponent ( $x^{-2}$ still means $1/x^2$ ). What actually licenses termwise integration is that each resulting exponent avoids the one excluded case, not that a denominator has vanished.
DIt cannot be — any fraction like this needs a technique beyond P1, such as logarithms.
Only a split term that is exactly $x^{-1}$ needs logarithms; that is the one excluded case, not a blanket rule about fractions in general. This particular split clears the exclusion on both terms.

Traps tested: Over generalizes the n equals minus 1 exclusion · Denominator removal misunderstood as the reason

Question 32 marks

Which of these, once split into separate powers of x, produces a term that needs a technique beyond what P1 covers?

A $\frac{4x^3 - 2}{x^2}$
This splits to $4x - 2x^{-2}$ — exponents 1 and $-2$ , neither the excluded $n=-1$ . This one stays entirely inside P1.
$\frac{6x - 3}{x}$
Correct — this is the one. $\frac{6x}{x} = 6$ and $\frac{-3}{x} = -3x^{-1}$ : the constant term in the numerator, divided by a denominator of exactly $x^1$ , lands exactly on the excluded case. $\int x^{-1}\,dx$ needs natural logarithms, which spec 5.2 explicitly excludes and P1 does not teach.
C $\frac{5x^4 + x^2}{x^2}$
This splits to $5x^2 + 1$ — exponents 2 and 0, both safe. A denominator power that exactly matches a numerator term's own power produces a plain constant, not a negative power at all.
D $\frac{9x^5 - 6x^3}{3x^3}$
This splits to $3x^2 - 2$ — exponents 2 and 0, both safe. Every exponent produced by a real WMA11-style fraction is safe except the one specific case where a numerator term's power exactly equals the denominator's power minus one.

Traps tested: Over generalizes the n equals minus 1 exclusion

Question 42 marks

$\frac{dy}{dx} = \frac{4x^3-9}{x^2}$ and the curve $y=f(x)$ passes through $(1,2)$ . A student splits and integrates correctly to reach $y = 2x^2 + 9x^{-1} + c$ , then substitutes the point to find $c=-9$ , giving the final answer $y = 2x^2 + 9x^{-1} - 9$ . What did this student do right that the research bank's own four-series evidence says many real candidates get wrong?

They wrote $+c$ the moment the integration itself was finished — before ever substituting the point — rather than leaving it out and hoping the final numeric answer would still come out right.
Correct. That exact step is independently confirmed across four series (Jan 2023 Q3, Oct 2021 Q1, Oct 2020 Q9, Jan 2025 Q6(b)) as the single most repeated "silly" mark loss on integration questions in the whole corpus researched for this course.
BNothing unusual — integrating $-9x^{-2}$ should actually give $-9x^{-1}$ , not $+9x^{-1}$ , so this working has an uncaught sign error.
This is a real trap in general, but it has not happened here: $\int -9x^{-2}\,dx = -9 \cdot \frac{x^{-1}}{-1} = 9x^{-1}$ — the two negative signs cancel. Suspecting a step that is actually correct is a wasted check, not a wrong one, but this particular suspicion is wrong.
CNothing unusual — writing c before substituting the point instead of after makes no difference, only to the order the working is written in.
It is a real difference, not a stylistic one. A candidate who never writes c separately at all, and instead tries to fold the point directly into an un-parameterised expression, has no free unknown left for the point to determine. Writing $+c$ immediately is what keeps that unknown available to be solved for.
DNothing unusual — since c turned out negative, the constant of integration was not really needed here.
The sign of the value c eventually takes has nothing to do with whether writing it was necessary. Spec 5.1 requires a constant of integration every time indefinite integration is performed, regardless of what value it later turns out to take.

Traps tested: Assumes sign error where none occurred · Deferred c treated as equivalent · Negative c value misread as optional

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2023 · Q3 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA11.

Splitting an Algebraic Fraction into Separate Terms Before Integrating

Why a fraction under an integral sign needs an extra step first

"Integrate the top, integrate the bottom, then divide" is not risky — it is provably a different calculation

Complete the chain — dydx=3x4−5x2\frac{dy}{dx} = \frac{3x^4 - 5}{x^2}dxdy​=x23x4−5​, curve through (1,10)(1, 10)(1,10): find y

Deep dive

Practice this for real

Complete the chain — $\frac{dy}{dx} = \frac{3x^4 - 5}{x^2}$ , curve through $(1, 10)$ : find y