11 min readPathway sheet3-mark questions, sometimes a 5-mark answer; also a JEE favourite

Integration by partial fractions

Split a rational function P(x)/Q(x) into simple fractions whose integrals are logs or tan⁻¹. Make it proper first, then match the factors of Q(x).

Rational functionP(x)/Q(x), where P and Q are polynomials and Q(x) ≠ 0

Proper fractionDegree of P(x) < degree of Q(x). If not, divide first: P/Q = quotient + remainder/Q.

Two integrals you need∫ dx/(x − a) = log|x − a| + C; ∫ dx/(x − a)² = −1/(x − a) + C

Fast constantsIn the identity, put x = each root of Q(x) to find its constant at once

Which form to write

Rational function	Partial fraction form
(px + q)/((x − a)(x − b)), a ≠ b	A/(x − a) + B/(x − b)
(px + q)/(x − a)²	A/(x − a) + B/(x − a)²
(px² + qx + r)/((x − a)(x − b)(x − c))	A/(x − a) + B/(x − b) + C/(x − c)
(px² + qx + r)/((x − a)²(x − b))	A/(x − a) + B/(x − a)² + C/(x − b)
(px² + qx + r)/((x − a)(x² + bx + c)), where x² + bx + c has no real factors	A/(x − a) + (Bx + C)/(x² + bx + c)

Method

1
Check that the fraction is proper. If not, divide the numerator by the denominator.
2
Factorise the denominator completely.
3
Write the partial fraction form from the table, with unknown constants A, B, C.
4
Multiply both sides by the denominator to get an identity.
5
Put x = each root to find constants quickly. Compare coefficients for any constant left.
6
Integrate each simple fraction and add one constant of integration.

Example 1: two linear factors

Problem

Find ∫ (5x + 1)/(x² + x − 2) dx.

Working

Factorise: x² + x − 2 = (x − 1)(x + 2).
Write (5x + 1)/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2).
Multiply out: 5x + 1 = A(x + 2) + B(x − 1).
Put x = 1: 6 = 3A, so A = 2.
Put x = −2: −9 = −3B, so B = 3.
Integral = 2 ∫ dx/(x − 1) + 3 ∫ dx/(x + 2).

Answer

2 log|x − 1| + 3 log|x + 2| + C

Example 2: a repeated factor

Problem

Find ∫ (x + 5)/((x − 1)²(x + 2)) dx.

Working

Form: A/(x − 1) + B/(x − 1)² + C/(x + 2).
Identity: x + 5 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)².
Put x = 1: 6 = 3B, so B = 2.
Put x = −2: 3 = 9C, so C = 1/3.
Coefficient of x²: 0 = A + C, so A = −1/3.
Integral = −(1/3) log|x − 1| + 2 × (−1/(x − 1)) + (1/3) log|x + 2|.

Answer

(1/3) log|(x + 2)/(x − 1)| − 2/(x − 1) + C

Example 3: an improper fraction

Problem

Find ∫ (x³ + 1)/(x² − 4) dx.

Working

Degree 3 is more than degree 2, so divide first: x³ + 1 = x(x² − 4) + (4x + 1).
So (x³ + 1)/(x² − 4) = x + (4x + 1)/((x − 2)(x + 2)).
Write (4x + 1)/((x − 2)(x + 2)) = A/(x − 2) + B/(x + 2), so 4x + 1 = A(x + 2) + B(x − 2).
Put x = 2: 9 = 4A, so A = 9/4. Put x = −2: −7 = −4B, so B = 7/4.
Integral = ∫ x dx + (9/4) ∫ dx/(x − 2) + (7/4) ∫ dx/(x + 2).

Answer

x²/2 + (9/4) log|x − 2| + (7/4) log|x + 2| + C

Common slips

Forgetting to divide when the fraction is improper.
Writing only B/(x − a)² for a repeated factor. You need both A/(x − a) and B/(x − a)².
Writing A/(x² + bx + c) instead of (Bx + C)/(x² + bx + c) for a quadratic factor.
Missing the modulus in log|x − a|.
For sec² x/((tan x + 1)(tan x + 3)), put t = tan x first, then use partial fractions in t.

In real life

The logistic model for the growth of bacteria in curd, or fish in a lake, leads to ∫ dP/(P(K − P)), where K is the largest population possible. It is solved by partial fractions.

Exam angle

2 marks: “Find ∫ dx/(x(x + 1)).” 1/(x(x + 1)) = 1/x − 1/(x + 1). Integral = log|x| − log|x + 1| + C = log|x/(x + 1)| + C.

3 marks: “Find ∫ sec² x/((tan x + 1)(tan x + 3)) dx.” Put t = tan x, dt = sec² x dx. 1/((t + 1)(t + 3)) = (1/2)[1/(t + 1) − 1/(t + 3)]. Answer: (1/2) log|(tan x + 1)/(tan x + 3)| + C.

5 marks: “Find ∫ (2x² + 3)/((x − 1)(x² + 4)) dx.” Write A/(x − 1) + (Bx + D)/(x² + 4), so 2x² + 3 = A(x² + 4) + (Bx + D)(x − 1). x = 1 gives A = 1; x² terms give A + B = 2, so B = 1; constants give 4A − D = 3, so D = 1. Answer: log|x − 1| + (1/2) log(x² + 4) + (1/2) tan⁻¹(x/2) + C.

Remember it

Proper, factorise, put x = roots. Then every piece is a log or a tan⁻¹.