Integration Using Partial Fractions

This lesson covers integrating rational functions using partial fraction decomposition — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to decompose fractions into simpler parts and integrate each part, often producing logarithmic results.

---

Why Partial Fractions?

A rational function is a fraction where both the numerator and denominator are polynomials. Many rational functions cannot be integrated directly, but when decomposed into partial fractions, each part can be integrated easily.

For example:

∫ 1/[(x - 1)(x + 2)] dx is difficult as it stands.

But 1/[(x - 1)(x + 2)] = (1/3)/(x - 1) - (1/3)/(x + 2)

Now each fraction can be integrated using the standard result ∫ 1/(x - a) dx = ln|x - a| + c.

---

Review of Partial Fraction Decomposition

Type 1: Distinct Linear Factors

If the denominator has distinct (different) linear factors:

f(x) / [(x - a)(x - b)] = A/(x - a) + B/(x - b)

Example: Express 5x + 1 / [(x + 1)(x - 2)] in partial fractions.

5x + 1 / [(x + 1)(x - 2)] = A/(x + 1) + B/(x - 2)

Multiply through by (x + 1)(x - 2): 5x + 1 = A(x - 2) + B(x + 1)

Substituting x = 2: 11 = 3B → B = 11/3 Substituting x = -1: -4 = -3A → A = 4/3

So: (5x + 1)/[(x + 1)(x - 2)] = (4/3)/(x + 1) + (11/3)/(x - 2)

Type 2: Repeated Linear Factor

If a factor is repeated:

f(x) / [(x - a)²(x - b)] = A/(x - a) + B/(x - a)² + C/(x - b)

Type 3: Three Distinct Linear Factors

f(x) / [(x - a)(x - b)(x - c)] = A/(x - a) + B/(x - b) + C/(x - c)

Exam Tip: Before decomposing, always check that the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first.

---

Integrating Partial Fractions

The Key Result

The integral of A/(x - a) is:

∫ A/(x - a) dx = A ln|x - a| + c

This is the fundamental result that makes partial fraction integration work.

The Key Result for Repeated Factors

∫ A/(x - a)² dx = -A/(x - a) + c

This uses the power rule (rewrite as A(x - a)⁻² and integrate).

---

Worked Examples

Example 1: Distinct Linear Factors

Evaluate ∫ (3x + 5) / [(x + 1)(x + 3)] dx

Step 1: Decompose into partial fractions. (3x + 5) / [(x + 1)(x + 3)] = A/(x + 1) + B/(x + 3)

Multiply through: 3x + 5 = A(x + 3) + B(x + 1)

x = -1: 2 = 2A → A = 1 x = -3: -4 = -2B → B = 2

So: (3x + 5)/[(x + 1)(x + 3)] = 1/(x + 1) + 2/(x + 3)

Step 2: Integrate each term. ∫ [1/(x + 1) + 2/(x + 3)] dx = ln|x + 1| + 2 ln|x + 3| + c

Example 2: Three Linear Factors

Evaluate ∫ (6x² + 5x - 2) / [x(x - 1)(2x + 1)] dx

Step 1: Decompose. (6x² + 5x - 2) / [x(x - 1)(2x + 1)] = A/x + B/(x - 1) + C/(2x + 1)

Multiply through: 6x² + 5x - 2 = A(x - 1)(2x + 1) + Bx(2x + 1) + Cx(x - 1)

x = 0: -2 = A(-1)(1) → A = 2 x = 1: 9 = B(1)(3) → B = 3 x = -1/2: 6(1/4) + 5(-1/2) - 2 = C(-1/2)(-3/2) 3/2 - 5/2 - 2 = (3/4)C → -3 = (3/4)C → C = -4

Step 2: Integrate. ∫ [2/x + 3/(x - 1) + (-4)/(2x + 1)] dx

= 2 ln|x| + 3 ln|x - 1| - 4 × (1/2) ln|2x + 1| + c

= 2 ln|x| + 3 ln|x - 1| - 2 ln|2x + 1| + c

Important: When integrating A/(ax + b), the result is (A/a) ln|ax + b| + c. Don't forget to divide by the coefficient of x.

Example 3: Repeated Factor

Evaluate ∫ (3x + 2) / (x + 1)² dx

Step 1: Decompose. (3x + 2) / (x + 1)² = A/(x + 1) + B/(x + 1)²

Multiply through: 3x + 2 = A(x + 1) + B

x = -1: -1 = B → B = -1 Comparing x coefficients: 3 = A → A = 3

Step 2: Integrate. ∫ [3/(x + 1) + (-1)/(x + 1)²] dx

= 3 ln|x + 1| + 1/(x + 1) + c

Example 4: Definite Integral with Partial Fractions

Evaluate ∫ from 2 to 4 of 3/[(x - 1)(x + 2)] dx

Step 1: Decompose. 3/[(x - 1)(x + 2)] = A/(x - 1) + B/(x + 2)

3 = A(x + 2) + B(x - 1)

x = 1: 3 = 3A → A = 1 x = -2: 3 = -3B → B = -1

Step 2: Integrate and evaluate. ∫ from 2 to 4 of [1/(x - 1) - 1/(x + 2)] dx

= [ln|x - 1| - ln|x + 2|] from 2 to 4

= [ln|x - 1| - ln|x + 2|] from 2 to 4

= (ln 3 - ln 6) - (ln 1 - ln 4)

= ln 3 - ln 6 - 0 + ln 4

= ln(3 × 4 / 6)

= ln 2

Exam Tip: Use logarithm laws to simplify your answer where possible. Remember: ln a + ln b = ln(ab) and ln a - ln b = ln(a/b).

---

Improper Fractions

If the degree of the numerator is equal to or greater than the degree of the denominator, you must perform polynomial long division first.

Example: Find ∫ (x² + 3x + 1)/(x + 2) dx

Divide: x² + 3x + 1 ÷ (x + 2) = x + 1 remainder -1

So: (x² + 3x + 1)/(x + 2) = x + 1 - 1/(x + 2)

Integrate: ∫ [x + 1 - 1/(x + 2)] dx = x²/2 + x - ln|x + 2| + c

---

Common Mistakes

1. Not dividing first — If the numerator degree ≥ denominator degree, you MUST divide first.
2. Forgetting the 1/a factor — When integrating 1/(ax + b), the answer is (1/a) ln|ax + b|, not just ln|ax + b|.
3. Missing modulus signs — Always write ln|...| not ln(...).
4. Arithmetic errors in finding constants — Double-check A, B, C by substituting a value back into the original equation.

---

Summary

• Decompose rational functions into partial fractions before integrating.
• The key integral: ∫ A/(x - a) dx = A ln|x - a| + c.
• For repeated factors (x - a)², include terms A/(x - a) and B/(x - a)².
• If the fraction is improper, perform long division first.
• Use log laws to simplify definite integral answers.
• Always include modulus signs inside logarithms.

A-Level Deep Dive: Integration Using Partial Fractions

Spec mapping

Edexcel 9MA0-02 specification section 10 — Integration covers using partial fractions that are linear in the denominator (refer to the official specification document for exact wording). This sub-strand sits in Year 2 Pure and is examined on Paper 2. It builds directly on section 4 (Algebra and functions, partial-fraction decomposition), which provides the algebraic prerequisite, and feeds section 11 (Differential equations, separable form), where partial fractions appear inside $\int 1/y(a-y)\,dy$∫1/y(a−y)dy logistic-type integrals. The Edexcel formula booklet lists $\int 1/x\,dx = \ln|x| + C$∫1/xdx=ln∣x∣+C but does not list the chain-extended form $\int 1/(ax+b)\,dx = (1/a)\ln|ax+b| + C$∫1/(ax+b)dx=(1/a)ln∣ax+b∣+C — it must be derived or memorised.

---

Worked example with full mark scheme

Question (8 marks): Find $\displaystyle\int \dfrac{3x + 5}{(x + 1)(x - 2)}\,dx$∫(x+1)(x−2)3x+5​dx, giving your answer in the form $p\ln|x+1| + q\ln|x-2| + C$pln∣x+1∣+qln∣x−2∣+C where $p$p and $q$q are constants to be found.

Solution with mark scheme:

Step 1 — set up the partial-fraction decomposition.

$\dfrac{3x + 5}{(x + 1)(x - 2)} = \dfrac{A}{x + 1} + \dfrac{B}{x - 2}$(x+1)(x−2)3x+5​=x+1A​+x−2B​

M1 — correct form of the partial-fraction split with two distinct linear factors. A frequent slip is to write $\dfrac{Ax + B}{(x+1)(x-2)}$(x+1)(x−2)Ax+B​, which is the same expression rearranged and earns nothing.

Step 2 — clear denominators.

$3x + 5 = A(x - 2) + B(x + 1)$3x+5=A(x−2)+B(x+1)

M1 — multiplying through by $(x+1)(x-2)$(x+1)(x−2) correctly.

Step 3 — solve for $A$A and $B$B by substitution.

Set $x = 2$x=2: $6 + 5 = 11 = B(3)$6+5=11=B(3), so $B = \tfrac{11}{3}$B=311​.

Set $x = -1$x=−1: $-3 + 5 = 2 = A(-3)$−3+5=2=A(−3), so $A = -\tfrac{2}{3}$A=−32​.

A1 — both constants correct. The cover-up technique (substituting the root of each linear factor) is the fastest route; equating coefficients of $x$x and the constant term is an acceptable alternative but slower.

Step 4 — rewrite the integrand.

$\dfrac{3x + 5}{(x+1)(x-2)} = -\dfrac{2/3}{x + 1} + \dfrac{11/3}{x - 2}$(x+1)(x−2)3x+5​=−x+12/3​+x−211/3​

A1 — correct decomposed form. This is the bridge from algebra to integration.

Step 5 — integrate term by term.

$\int \left( -\dfrac{2/3}{x+1} + \dfrac{11/3}{x-2} \right) dx = -\dfrac{2}{3}\ln|x+1| + \dfrac{11}{3}\ln|x-2| + C$∫(−x+12/3​+x−211/3​)dx=−32​ln∣x+1∣+311​ln∣x−2∣+C

M1 — applying $\int 1/(ax+b)\,dx = (1/a)\ln|ax+b| + C$∫1/(ax+b)dx=(1/a)ln∣ax+b∣+C to each term. Here $a = 1$a=1 in both, so the $1/a$1/a factor is invisible — but the modulus signs are required.

A1 — correct first term $-\tfrac{2}{3}\ln|x+1|$−32​ln∣x+1∣.

A1 — correct second term $\tfrac{11}{3}\ln|x-2|$311​ln∣x−2∣.

B1 — constant of integration $+C$+C included. Edexcel routinely awards a separate B mark for $+C$+C on indefinite integrals; omitting it costs a guaranteed mark.

Total: 8 marks (M3 A4 B1). Final form: $p = -\tfrac{2}{3}$p=−32​, $q = \tfrac{11}{3}$q=311​.

---

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): Find $\displaystyle\int_0^1 \dfrac{4}{(2x + 1)(x + 3)}\,dx$∫01​(2x+1)(x+3)4​dx, giving your answer in the form $\ln k$lnk where $k$k is a rational number to be found.

Mark scheme decomposition by AO:

• M1 (AO1.1a) — correct partial-fraction setup $\dfrac{4}{(2x+1)(x+3)} = \dfrac{A}{2x+1} + \dfrac{B}{x+3}$(2x+1)(x+3)4​=2x+1A​+x+3B​.
• A1 (AO1.1b) — correct constants $A = \tfrac{8}{5}$A=58​, $B = -\tfrac{4}{5}$B=−54​ (cover-up: $x = -\tfrac{1}{2}$x=−21​ gives $4 = A \cdot \tfrac{5}{2}$4=A⋅25​; $x = -3$x=−3 gives $4 = B(-5)$4=B(−5)).
• M1 (AO1.1b) — integrating, including the $1/a$1/a factor for the $2x+1$2x+1 term: $\tfrac{8}{5} \cdot \tfrac{1}{2}\ln|2x+1| - \tfrac{4}{5}\ln|x+3| = \tfrac{4}{5}\ln|2x+1| - \tfrac{4}{5}\ln|x+3|$58​⋅21​ln∣2x+1∣−54​ln∣x+3∣=54​ln∣2x+1∣−54​ln∣x+3∣.
• M1 (AO1.1b) — applying limits $0$0 to $1$1: $\tfrac{4}{5}[\ln 3 - \ln 4] - \tfrac{4}{5}[\ln 4 - \ln 3] = \tfrac{8}{5}\ln(3/4)$54​[ln3−ln4]−54​[ln4−ln3]=58​ln(3/4).
• A1 (AO2.5) — combining using log laws: $\tfrac{8}{5}\ln(3/4) = \ln((3/4)^{8/5})$58​ln(3/4)=ln((3/4)8/5), so $k = (3/4)^{8/5}$k=(3/4)8/5.
• A1 (AO2.5) — final exact rational presentation, typically left as $\tfrac{8}{5}\ln(3/4)$58​ln(3/4) or equivalent.

Total: 6 marks split AO1 = 4, AO2 = 2. AO2 marks reward the log-law manipulation that compresses two log differences into a single $\ln k$lnk form.

---

Synoptic links

Connects to:

1. Section 4 — Algebra and functions (partial-fraction decomposition): the algebraic prerequisite. Without confidence decomposing $\dfrac{3x+5}{(x+1)(x-2)}$(x+1)(x−2)3x+5​ into $\dfrac{A}{x+1} + \dfrac{B}{x-2}$x+1A​+x−2B​, the integration step is unreachable. Cover-up technique is shared exactly.

2. Section 6 — Exponentials and logarithms: the integration result is a sum of natural logs, so log laws ($\ln a + \ln b = \ln(ab)$lna+lnb=ln(ab), $\ln a - \ln b = \ln(a/b)$lna−lnb=ln(a/b), $n\ln a = \ln a^n$nlna=lnan) are needed to compress the answer into the $\ln k$lnk form Edexcel often demands.

3. Section 11 — Differential equations (separable form): logistic-type ODEs $\dfrac{dy}{dt} = ky(N - y)$dtdy​=ky(N−y) separate to $\int \dfrac{1}{y(N - y)}\,dy = \int k\,dt$∫y(N−y)1​dy=∫kdt, and the left-hand integral is exactly a partial-fractions integration. Population dynamics, chemical kinetics and epidemic models all hit this pattern.

4. Section 4 — Binomial expansion: an alternative route to $\dfrac{1}{(1+x)(1-2x)}$(1+x)(1−2x)1​-type integrands is to expand each linear factor as a binomial series for $|x| < 1/2$∣x∣<1/2 and integrate term by term. This trades an exact log answer for a power-series approximation — a synoptic comparison Edexcel sometimes invites in extended questions.