Integration by partial fractions
NCERT Class 12 Mathematics Chapter 7 Section 7.5
Quick answer
To integrate a rational function, first divide if it is improper, so that the remaining fraction has a numerator of lower degree than its denominator. Then factorise the denominator and split the fraction into partial fractions: A/(x − a) for each linear factor, A/(x − a) + B/(x − a) for a repeated one, and (Bx + C)/(x + bx + c) for a quadratic that cannot be factorised. Each piece integrates to a logarithm, a power or an inverse tangent.
With your textbook
NCERT Mathematics Part II (Class 12), Chapter 7 “Integrals”, Section 7.5 “Integration by Partial Fractions”, pages 252–259.
Open the official NCERT chapter (PDF, 5.7 MB, ncert.nic.in) (opens in a new tab)
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What is a partial fraction decomposition?
Partial fractions
It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition.
NCERT Mathematics Part II (Class 12), Section 7.5, page 253
- A rational function P(x)/Q(x) is proper if the degree of P(x) is less than the degree of Q(x), and improper otherwise (NCERT page 252).
- An improper one is divided first: P(x)/Q(x) = T(x) + P(x)/Q(x), where T(x) is a polynomial and P(x)/Q(x) is proper (NCERT page 253).
In plain words: a rational function is a polynomial divided by a polynomial. Once it is proper, its denominator is factorised and the fraction is broken into small pieces such as A/(x − a), each of which integrates to a logarithm, a power or an inverse tangent. This reverses adding fractions over a common denominator.
The forms of the partial fractions
NCERT Table 7.2 (page 253) gives the shape of the pieces for each kind of denominator. A, B and C are real numbers to be found. In the three distinct linear-factor row, a, b, c are pairwise distinct; in the repeated-factor row, a ≠ b. The general form includes every indicated power even though some coefficients may be zero.
- (px + q)/((x − a)(x − b)), with 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 cannot be factorised further: A/(x − a) + (Bx + C)/(x2 + bx + c).
Finding A, B and C. Multiply both sides by the denominator and compare the coefficients of each power of x, as NCERT does. NCERT’s Remark (page 254) adds that the result is an identity, true for every permissible x. Once the denominators are cleared, as in 1 = A(x + 2) + B(x + 1), both sides are polynomials that agree at every permissible x, so they agree at every x (reasoned from NCERT’s Remark). So you may also put in convenient values, even x = a, which removes every term except the one with A.
Distinct and repeated linear factors
NCERT Example 11 (page 254): 1/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2) gives 1 = A(x + 2) + B(x + 1), so A + B = 0 and 2A + B = 1, that is A = 1 and B = −1:
∫dx/((x + 1)(x + 2)) = log |(x + 1)/(x + 2)| + C.
An improper fraction is divided first. NCERT Example 12 (pages 254–255): (x2 + 1)/(x2 − 5x + 6) = 1 + (5x − 5)/((x − 2)(x − 3)), and the proper part splits as −5/(x − 2) + 10/(x − 3). So the integral is x − 5 log |x − 2| + 10 log |x − 3| + C.
A repeated factor needs every power. NCERT Example 13 (page 255): (3x − 2)/((x + 1)2(x + 3)) = A/(x + 1) + B/(x + 1)2 + C/(x + 3), with A = 11/4, B = −5/2 and C = −11/4. The middle term integrates to a power, not a logarithm:
∫(3x − 2)/((x + 1)2(x + 3)) dx
= (11/4) log |(x + 1)/(x + 3)| + 5/(2(x + 1)) + C.
A quadratic factor
When a quadratic factor cannot be factorised, its piece has a linear top, Bx + C. NCERT Example 16 (pages 257–258): (x2 + x + 1)/((x + 2)(x2 + 1)) = A/(x + 2) + (Bx + C)/(x2 + 1), with A = 3/5, B = 2/5 and C = 1/5. Writing the second piece as (1/5)(2x + 1)/(x2 + 1):
∫(x2 + x + 1)/((x + 2)(x2 + 1)) dx
= (3/5) log |x + 2| + (1/5) log |x2 + 1| + (1/5) tan x + C.
NCERT slip (page 257): Example 16 refers to Table 2.2(5), in square brackets; the table is Table 7.2, row 5, on page 253. It does not change the working.
A substitution can come first. NCERT Example 14 (page 256) puts y = x2 only to find the fractions: x2/((x2 + 1)(x2 + 4)) = −1/(3(x2 + 1)) + 4/(3(x2 + 4)), which integrates to −(1/3) tan x + (2/3) tan (x/2) + C. NCERT Example 15 (pages 256–257) puts y = sin φ, which turns the integral into ∫(3y − 2)/(y − 2)2 dy; the answer is 3 log (2 − sin φ) + 4/(2 − sin φ) + C.
Solved examples
Example 1: Three distinct factors (NCERT Exercise 7.5, Q3, page 258)
Find ∫(3x − 1)/
Write the integrand as A/(x − 1) + B/(x − 2) + C/(x − 3), so 3x − 1 = A(x − 2)(x − 3) + B(x − 1)(x − 3) + C(x − 1)(x − 2). Putting x = 1, 2, 3 in turn: 2 = 2A, 5 = −B and 8 = 2C, so A = 1, B = −5 and C = 4.
Answer: log |x − 1| − 5 log |x − 2| + 4 log |x − 3| + C.
Example 2: Divide first (NCERT Exercise 7.5, Q12, page 258)
Find ∫(x3 + x + 1)/(x2 − 1) dx.
The fraction is improper: x3 + x + 1 = x(x2 − 1) + 2x + 1, so the integrand is x + (2x + 1)/((x − 1)(x + 1)). The proper part is A/(x − 1) + B/(x + 1) with A = 3/2 (put x = 1) and B = 1/2 (put x = −1).
Answer: x2/2 + (3/2) log |x − 1| + ½ log |x + 1| + C.
Example 3: A repeated factor (NCERT Exercise 7.5, Q8, page 258)
Find ∫x/((x − 1)2(x + 2)) dx.
Write x = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)2. x = 1 gives B = 1/3; x = −2 gives C = −2/9; the coefficient of x2 gives A + C = 0, so A = 2/9. The B term integrates to −B/(x − 1).
Answer: (2/9) log |(x − 1)/(x + 2)| − 1/(3(x − 1)) + C.
Example 4: A quadratic factor (NCERT Exercise 7.5, Q13, page 258)
Find ∫2/((1 − x)(1 + x2)) dx.
Write 2 = A(1 + x2) + (Bx + C)(1 − x). x = 1 gives A = 1. Comparing the x2 terms, A − B = 0, so B = 1; comparing the constants, A + C = 2, so C = 1. The pieces 1/(1 − x), x/(1 + x2) and 1/(1 + x2) give −log |1 − x|, ½ log (1 + x2) and tan x.
Answer: −log |1 − x| + ½ log (1 + x2) + tan x + C.
Example 5: Choose the correct answer (NCERT Exercise 7.5, Q23, page 259)
∫dx/(x(x2 + 1)) equals (A) log |x| − ½ log (x2 + 1) + C; (B) log |x| + ½ log (x2 + 1) + C; (C) −log |x| + ½ log (x2 + 1) + C; (D) ½ log |x| + log (x2 + 1) + C.
1/(x(x2 + 1)) = A/x + (Bx + C)/(x2 + 1) gives A = 1, B = −1 and C = 0, so the integrand is 1/x − x/(x2 + 1).
Answer: (A).
Common mistakes to avoid
Common mistake
Splitting an improper fraction
(x2 + 1)/(x2 − 5x + 6) is not A/(x − 2) + B/(x − 3): the top has the same degree as the bottom. Divide first, as in NCERT Example 12, and split only the proper part.
Common mistake
Leaving out a power of a repeated factor
(x + 1)2 in the denominator needs both A/(x + 1) and B/(x + 1)2. With only one of them, the coefficients cannot be matched.
Common mistake
A constant over a quadratic factor
An unfactorisable x2 + bx + c needs (Bx + C)/(x2 + bx + c), not just B/(x2 + bx + c). In Example 5 the piece is −x/(x2 + 1), which a constant top could never give.
Quick revision summary
- Divide an improper fraction first; then factorise the denominator of the proper part.
- Each linear factor (x − a) gives A/(x − a); a repeated factor (x − a)2 gives A/(x − a) + B/(x − a)2; an unfactorisable quadratic gives (Bx + C)/(x2 + bx + c).
- Find the constants by comparing coefficients, or by putting convenient values of x; then integrate each piece.
Formula sheet
- A/(x − a) → A log |x − a|
- B/(x − a) → −B/(x − a)
- (px + q)/(x + 1) → (p/2) log (x + 1) + q tan x
Questions students ask
Why does partial fraction decomposition work?
NCERT states that every proper rational function admits a decomposition of the prescribed form (page 253). Counting unknowns explains the form but alone does not prove that the equations are solvable: pieces of the kinds in NCERT’s Table 7.2 bring in exactly as many unknown constants as the degree of the denominator (reasoned): one for each linear factor, two for a squared one and two, B and C, for an unfactorisable quadratic. Clearing the denominator then gives a polynomial identity whose coefficients, compared power by power, give just that many equations, and NCERT states that such a split is always possible (page 253). For a denominator (x − 1)(x + 2)2(x2 + 3), of degree 5, the count is 1 + 2 + 2 = 5 unknowns.
What is the shortcut, or cover-up rule, for finding partial fractions?
The shortcut often called the cover-up rule (beyond the NCERT text, reasoned from putting x = a) gives the constant over a distinct linear factor x − a at once: cover that factor in the original fraction and put x = a in what is left. For (x + 7)/((x − 1)(x + 3)), covering x − 1 and putting x = 1 gives 8/4 = 2, and covering x + 3 and putting x = −3 gives 4/(−4) = −1, so the fraction is 2/(x − 1) − 1/(x + 3), and its integral is 2 log |x − 1| minus log |x + 3|, plus C. For a repeated factor (x − a)2 the rule gives only the constant over the square, and for a quadratic factor nothing; find the rest by comparing coefficients.
When should you use partial fractions to integrate?
Use partial fractions for a rational function (a polynomial over a polynomial) whose denominator factorises, once quicker routes are ruled out: a top that is a multiple of the derivative of the bottom gives a logarithm by substitution, and a single quadratic below may fit a particular integral. So ∫(2x + 5)/(x2 + 5x + 4) dx needs no partial fractions, though the bottom is (x + 1)(x + 4): its top is the derivative of the bottom, and the integral is log |x2 + 5x + 4| + C. A substitution can also make an integrand rational first, as y = sin φ does in NCERT Example 15.
How do you write partial fractions for (x − a)3 or a repeated quadratic factor?
Beyond the NCERT text (reasoned), a factor repeated three times, (x − a)3, needs one term per power, A/(x − a), B/(x − a)2 and C/(x − a)3, and a repeated unfactorisable quadratic such as (x2 + 1)2 needs a linear top over each power; NCERT’s Table 7.2 (page 253) stops at a squared linear factor and a single quadratic one. For x2/(x − 1)3, writing x2 as (x − 1)2 + 2(x − 1) + 1 splits it into 1/(x − 1), 2/(x − 1)2 and 1/(x − 1)3, so the integral is log |x − 1| − 2/(x − 1), minus 1/(2(x − 1)2), plus C.
Practice questions
Try each question first, then open the answer to check your working.
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Q1. Find ∫x/((x + 1)(x + 2)) dx (NCERT Exercise 7.5, Q1, page 258).
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x = A(x + 2) + B(x + 1): x = −1 gives A = −1, x = −2 gives B = 2. The integral is 2 log |x + 2| − log |x + 1| + C, that is log ((x + 2)2/|x + 1|) + C.
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Q2. Find ∫dx/(x2 − 9) (NCERT Exercise 7.5, Q2, page 258).
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1/((x − 3)(x + 3)) = (1/6)[1/(x − 3) − 1/(x + 3)], so the integral is (1/6) log |(x − 3)/(x + 3)| + C.
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Q3. Find ∫(3x − 1)/(x + 2)2 dx (NCERT Exercise 7.5, Q14, page 258).
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3x − 1 = 3(x + 2) − 7, so the integrand is 3/(x + 2) − 7/(x + 2)2, and the integral is 3 log |x + 2| + 7/(x + 2) + C.
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Q4. ∫x dx/((x − 1)(x − 2)) equals (A) log |(x − 1)2/(x − 2)| + C; (B) log |(x − 2)2/(x − 1)| + C; (C) log |((x − 1)/(x − 2))2| + C; (D) log |(x − 1)(x − 2)| + C (NCERT Exercise 7.5, Q22, page 259).
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x/((x − 1)(x − 2)) = −1/(x − 1) + 2/(x − 2), so the integral is 2 log |x − 2| − log |x − 1| = log |(x − 2)2/(x − 1)| + C: (B).