MathsISC Class 12 · Integrals

Integration by Partial Fractions: Every Case with Worked Examples

Updated 15 October 2026 · Fact-checked

To integrate a rational function, first make it proper by dividing if the numerator's degree is not less than the denominator's. Then split it into partial fractions: A/(x − a) for each linear factor, extra terms B/(x − a)² for a repeated factor, and (Bx + C)/(x² + bx + c) for an irreducible quadratic. Integrate each term separately.

Understand Integration by Partial Fractions

A rational function is a quotient of polynomials, $\frac{f ( x )}{g ( x )}$ . Such integrands appear constantly, and the general method is to break them into simple pieces whose integrals are logarithms, powers and inverse tangents.

Step 0: proper or improper? If $de g f \geq de g g$ , the fraction is improper. Divide first (long division) to write it as a polynomial plus a proper fraction. For example, $\frac{x ^{2} + 1}{x ^{2} - 5 x + 6} = 1 + \frac{5 x - 5}{x ^{2} - 5 x + 6}$ . The CISCE syllabus includes this case explicitly.

The forms of partial fractions (for a proper fraction), as listed in the syllabus:

Distinct linear factors: $\frac{x + 2}{( x - 3 ) ( x + 1 )} = \frac{A}{x - 3} + \frac{B}{x + 1}$ .
Repeated linear factor: $\frac{f ( x )}{( x - a ) ^{2} ( x - b )} = \frac{A}{x - a} + \frac{B}{( x - a ) ^{2}} + \frac{C}{x - b}$ . A factor squared needs two terms.
Irreducible quadratic factor (no real roots, like $x^{2} + 1$ ): $\frac{f ( x )}{( x - a ) ( x ^{2} + b x + c )} = \frac{A}{x - a} + \frac{B x + C}{x ^{2} + b x + c}$ . The numerator over the quadratic must be linear, $B x + C$ , not just a constant.

Finding the constants. Multiply both sides by the denominator to get a polynomial identity, then:

Substitute roots. Putting $x = a$ kills every term except the one with $A$ . This gives most constants at once.
Compare coefficients for the constants that substitution cannot reach (in repeated or quadratic cases), usually the coefficient of the highest power and the constant term.
Cover-up shortcut for a distinct linear factor: to find $A$ in $\frac{A}{x - a}$ , cover up $(x - a)$ in the original fraction and put $x = a$ in what is left.

The ISC 2025 examiners reported candidates who could not split into partial fractions correctly, so check your constants by putting in one extra value of $x$ .

Integrating the pieces.

$\int \frac{A}{x - a} d x = A lo g ∣ x - a ∣$
$\int \frac{B}{( x - a ) ^{2}} d x = - \frac{B}{x - a}$
$\int \frac{B x + C}{x ^{2} + k ^{2}} d x = \frac{B}{2} lo g (x^{2} + k^{2}) + \frac{C}{k} tan^{- 1} \frac{x}{k}$

Combine logarithms at the end if you like, using $lo g a - lo g b = lo g \frac{a}{b}$ .

Substitution first. Some integrands become rational only after a substitution. With $t = x^{2}$ , $\frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 3 )}$ becomes $\frac{1}{( t + 1 ) ( t + 3 )}$ . With $t = x^{4}$ , $\frac{1}{x ( x ^{4} + 1 )} = \frac{x ^{3}}{x ^{4} ( x ^{4} + 1 )}$ becomes $\frac{1}{4} \cdot \frac{1}{t ( t + 1 )}$ . With $t = e^{x}$ , $\frac{1}{e ^{x} - 1}$ becomes $\frac{1}{t ( t - 1 )}$ . Spotting these saves a lot of algebra.

Integration by Partial Fractions: formulas to remember

Distinct linear factors: $\frac{f ( x )}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}$; A = f(a)/(a − b) by the cover-up rule.
Repeated linear factor: $\frac{f ( x )}{( x - a ) ^{2} ( x - b )} = \frac{A}{x - a} + \frac{B}{( x - a ) ^{2}} + \frac{C}{x - b}$; Two terms for the squared factor.
Quadratic factor: $\frac{f ( x )}{( x - a ) ( x ^{2} + b x + c )} = \frac{A}{x - a} + \frac{B x + C}{x ^{2} + b x + c}$; Linear numerator over an irreducible quadratic.
Improper fraction: $de g f \geq de g g \Rightarrow \frac{f}{g} = q (x) + \frac{r ( x )}{g ( x )}$; Divide first.
Log integral: $\int \frac{d x}{x - a} = lo g ∣ x - a ∣ + C$; Keep the modulus.
Repeated term: $\int \frac{d x}{( x - a ) ^{2}} = - \frac{1}{x - a} + C$; From the power rule with power −2.
Standard pair: $\int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} lo g \frac{x - a}{x + a} + C$; A ready-made partial-fraction result.

How to solve Integration by Partial Fractions questions

Work through these steps for any rational integrand.

1Compare degrees. If the numerator's degree is at least the denominator's, divide and keep the quotient to integrate separately.
2Factorise the denominator completely into linear factors and irreducible quadratics.
3Write the partial-fraction form with unknown constants, one term per linear factor, two for a squared factor, and $\frac{B x + C}{\dots}$ for a quadratic.
4Multiply through by the denominator. Substitute the roots to find as many constants as possible.
5Find any remaining constants by comparing coefficients (highest power and constant term).
6Check by substituting one more value of $x$ into the identity.
7Integrate each term and add a single $+ C$ ; simplify logarithms if asked.

Integration by Partial Fractions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Partial fractions for a total time

Suppose a hiker climbing a steep trail slows down as she tires, so that the time she needs per kilometre at a distance $x$ km from the start is $\frac{x + 3}{( x + 1 ) ( x + 2 )}$ hours per km.

Find the time she takes to walk the first 2 km.

Try it first, then show the solutionSolution

Time $= \int_{0}^{2} \frac{x + 3}{( x + 1 ) ( x + 2 )} d x$ .
Partial fractions: $x + 3 = A (x + 2) + B (x + 1)$ . $x = - 1$ : $A = 2$ . $x = - 2$ : $B = - 1$ .
Time $= [2 lo g (x + 1) - lo g (x + 2)]_{0}^{2} = (2 lo g 3 - lo g 4) - (0 - lo g 2) = lo g \frac{9}{2}$ .
$lo g 4.5 \approx 1.50$ hours.

Answer: About 1.5 hours (exactly $lo g \frac{9}{2}$ hours).

Memory hook: Split the fraction, add the logs: a tiring climb in two log terms.

Partial fractions with a simple model

Imagine a water tank whose level falls more and more slowly. In a simple model, the time (in hours) for the level to fall from 4 m to 1 m is $T = \int_{1}^{4} \frac{6}{h ( h + 1 )} d h$ .

Find $T$ .

Try it first, then show the solutionSolution

$\frac{6}{h ( h + 1 )} = \frac{6}{h} - \frac{6}{h + 1}$ (cover-up: $A = 6$ , $B = - 6$ ).
$T = 6 [lo g \frac{h}{h + 1}]_{1}^{4} = 6 (lo g \frac{4}{5} - lo g \frac{1}{2}) = 6 lo g \frac{8}{5}$ .
$6 lo g 1.6 \approx 2.82$ hours.

Answer: About 2.82 hours ( $6 lo g \frac{8}{5}$ ).

Memory hook: 1/(h(h + 1)) = 1/h − 1/(h + 1): the simplest split of all.

Tricks to remember Integration by Partial Fractions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Divide before you split
Top-heavy fraction? Long division first, partial fractions second.
Use it when: Improper rational functions.
Squared means two
A squared factor (x − a)² needs A/(x − a) and B/(x − a)².
Use it when: Repeated linear factors.
Quadratic gets a line
Over an irreducible quadratic, the numerator is Bx + C, a line, not a constant.
Use it when: Avoiding missing terms.
Cover up
Hide (x − a), put x = a in the rest: that is A.
Use it when: Distinct linear factors in seconds.

Integration by Partial Fractions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Two distinct factors · 1 mark · MCQBoard level

$\int \frac{d x}{( x - 1 ) ( x - 2 )}$ equals:
(a) $lo g \frac{x - 1}{x - 2} + C$
(b) $lo g \frac{x - 2}{x - 1} + C$
(c) $lo g ∣ (x - 1) (x - 2) ∣ + C$
(d) $\frac{1}{( x - 1 ) ( x - 2 )} + C$

Try it first, then show the solutionSolution

$\frac{1}{( x - 1 ) ( x - 2 )} = \frac{- 1}{x - 1} + \frac{1}{x - 2}$ , so the integral is $lo g ∣ x - 2∣ - lo g ∣ x - 1∣$ .

Answer: (b).

Example 2: The syllabus example · 3 marks · Short answerBoard level

Evaluate $\int \frac{x + 4}{( x - 2 ) ( x + 1 )} d x$ .

Try it first, then show the solutionSolution

$\frac{x + 4}{( x - 2 ) ( x + 1 )} = \frac{A}{x - 2} + \frac{B}{x + 1}$ , so $x + 4 = A (x + 1) + B (x - 2)$ . (1 mark)
$x = 2$ : $6 = 3 A \Rightarrow A = 2$ . $x = - 1$ : $3 = - 3 B \Rightarrow B = - 1$ . (1 mark)
Integral $= 2 lo g ∣ x - 2∣ - lo g ∣ x + 1∣ + C = lo g \frac{( x - 2 ) ^{2}}{∣ x + 1∣} + C$ . (1 mark)

Answer: $2 lo g ∣ x - 2∣ - lo g ∣ x + 1∣ + C$ .

Example 3: A repeated linear factor · 3 marks · Short answerHarder board level

Evaluate $\int \frac{x}{( x - 1 ) ^{2} ( x + 2 )} d x$ .

Try it first, then show the solutionSolution

$x = A (x - 1) (x + 2) + B (x + 2) + C (x - 1)^{2}$ . (½ mark)
$x = 1$ : $1 = 3 B \Rightarrow B = \frac{1}{3}$ . $x = - 2$ : $- 2 = 9 C \Rightarrow C = - \frac{2}{9}$ . (1 mark)
Coefficient of $x^{2}$ : $0 = A + C \Rightarrow A = \frac{2}{9}$ . (½ mark)
Integral $= \frac{2}{9} lo g ∣ x - 1∣ - \frac{1}{3 ( x - 1 )} - \frac{2}{9} lo g ∣ x + 2∣ + C$ . (1 mark)

Answer: $\frac{2}{9} lo g \frac{x - 1}{x + 2} - \frac{1}{3 ( x - 1 )} + C$ .

Example 4: An improper fraction · 3 marks · Short answerHarder board level

Evaluate $\int \frac{x ^{2} + 1}{x ^{2} - 5 x + 6} d x$ .

Try it first, then show the solutionSolution

Divide: $\frac{x ^{2} + 1}{x ^{2} - 5 x + 6} = 1 + \frac{5 x - 5}{( x - 2 ) ( x - 3 )}$ . (1 mark)
$5 x - 5 = A (x - 3) + B (x - 2)$ : $x = 2$ gives $A = - 5$ ; $x = 3$ gives $B = 10$ . (1 mark)
Integral $= x - 5 lo g ∣ x - 2∣ + 10 lo g ∣ x - 3∣ + C$ . (1 mark)

Answer: $x - 5 lo g ∣ x - 2∣ + 10 lo g ∣ x - 3∣ + C$ .

Example 5: An irreducible quadratic factor · 3 marks · Short answerHarder board level

Evaluate $\int \frac{d x}{( x + 1 ) ( x ^{2} + 1 )}$ .

Try it first, then show the solutionSolution

$1 = A (x^{2} + 1) + (B x + C) (x + 1)$ . $x = - 1$ : $A = \frac{1}{2}$ . (1 mark)
$x^{2}$ : $0 = A + B \Rightarrow B = - \frac{1}{2}$ . Constant: $1 = A + C \Rightarrow C = \frac{1}{2}$ . (1 mark)
Integral $= \frac{1}{2} lo g ∣ x + 1∣ - \frac{1}{4} lo g (x^{2} + 1) + \frac{1}{2} tan^{- 1} x + C$ . (1 mark)

Answer: $\frac{1}{2} lo g ∣ x + 1∣ - \frac{1}{4} lo g (x^{2} + 1) + \frac{1}{2} tan^{- 1} x + C$ .

Example 6: Substitution before partial fractions · 5 marks · Long answerHOTS / competency

Evaluate $\int \frac{d x}{x ( x ^{4} + 1 )}$ and hence $\int_{1}^{2} \frac{d x}{x ( x ^{4} + 1 )}$ .

Try it first, then show the solutionSolution

Multiply top and bottom by $x^{3}$ : $\int \frac{x ^{3} d x}{x ^{4} ( x ^{4} + 1 )}$ . Put $t = x^{4}$ , $d t = 4 x^{3} d x$ . (1 mark)
The integral becomes $\frac{1}{4} \int \frac{d t}{t ( t + 1 )}$ . (1 mark)
$\frac{1}{t ( t + 1 )} = \frac{1}{t} - \frac{1}{t + 1}$ , so we get $\frac{1}{4} lo g \frac{t}{t + 1} = \frac{1}{4} lo g \frac{x ^{4}}{x ^{4} + 1} + C$ . (2 marks)
With limits: $\frac{1}{4} [lo g \frac{16}{17} - lo g \frac{1}{2}] = \frac{1}{4} lo g \frac{32}{17}$ . (1 mark)

Answer: $\frac{1}{4} lo g \frac{x ^{4}}{x ^{4} + 1} + C$ ; the definite integral is $\frac{1}{4} lo g \frac{32}{17}$ .

Where students lose marks in Integration by Partial Fractions

The slips examiners see most often, and what to write instead.

Mistake: Wrong form of partial fractions
The ISC 2025 examiners noted candidates who could not split into partial fractions correctly. The usual slips are missing a term for a repeated factor and putting a constant, instead of a linear numerator, over a quadratic.
Wrong working
$\frac{x ^{2}}{( x - 1 ) ^{2} ( x ^{2} + 1 )} = \frac{A}{( x - 1 ) ^{2}} + \frac{B}{x ^{2} + 1}$ .
Right working
$\frac{A}{x - 1} + \frac{B}{( x - 1 ) ^{2}} + \frac{C x + D}{x ^{2} + 1}$ .
Do this: Count unknowns: their number must equal the degree of the denominator.
Usually costs: Most of a 6-mark question.
Mistake: Skipping long division
Partial fractions only work for proper fractions.
Do this: Compare degrees before anything else.
Usually costs: 2 to 3 marks.
Mistake: Losing a coefficient when integrating
Terms like 1/(2x + 1) integrate to ½ log|2x + 1|; the ½ is often dropped.
Do this: Write the substitution factor whenever the linear term is not just x − a.
Usually costs: 1 mark.
Mistake: Failing on the basic integrals afterwards
The 2025 examiners also noted candidates who split correctly but then integrated the pieces wrongly.
Do this: Learn the three results: log for 1/(x − a), −1/(x − a) for 1/(x − a)², and log plus tan⁻¹ for (Bx + C)/(x² + k²).
Usually costs: 1 to 2 marks.

Exam tips

Write the identity after clearing denominators on a separate line; examiners give a method mark for it.
Show the values of $x$ you substituted, for example "put $x = 3$ : $5 = 4 A$ ".
Check the constants with one extra value of $x$ before integrating; it takes seconds and saves the whole question.
Keep moduli in logarithms of linear factors; $x^{2} + 1$ does not need one.
If a substitution such as $t = x^{2}$ or $t = e^{x}$ makes the integrand rational, say so clearly and change $d x$ correctly.

Practice questions on Integration by Partial Fractions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The partial fractions of $\frac{1}{x ^{2} - 1}$ are:
(a) $\frac{1}{x - 1} - \frac{1}{x + 1}$
(b) $\frac{1}{2} (\frac{1}{x - 1} - \frac{1}{x + 1})$
(c) $\frac{1}{2} (\frac{1}{x + 1} - \frac{1}{x - 1})$
(d) $\frac{1}{x - 1} + \frac{1}{x + 1}$
Hint
Cover-up at $x = 1$ and $x = - 1$ .
Answer
(b).
Q2 · 3 marks · Short answerBoard level
Evaluate $\int \frac{3 x - 1}{( x - 1 ) ( x - 2 ) ( x - 3 )} d x$ .
Hint
Cover-up for each factor.
Answer
$A = 1$ , $B = - 5$ , $C = 4$ : $lo g ∣ x - 1∣ - 5 lo g ∣ x - 2∣ + 4 lo g ∣ x - 3∣ + C$ .
Q3 · 3 marks · Short answerHarder board level
Evaluate $\int \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 3 )} d x$ .
Hint
Put $t = x^{2}$ .
Answer
$\int \frac{d t}{( t + 1 ) ( t + 3 )} = \frac{1}{2} lo g \frac{t + 1}{t + 3} = \frac{1}{2} lo g \frac{x ^{2} + 1}{x ^{2} + 3} + C$ .
Q4 · 3 marks · Short answerHarder board level
Evaluate $\int \frac{x ^{2} + x + 1}{( x + 2 ) ( x ^{2} + 1 )} d x$ .
Hint
Find $A$ from $x = - 2$ , then compare coefficients.
Answer
$A = \frac{3}{5}$ , $B = \frac{2}{5}$ , $C = \frac{1}{5}$ : $\frac{3}{5} lo g ∣ x + 2∣ + \frac{1}{5} lo g (x^{2} + 1) + \frac{1}{5} tan^{- 1} x + C$ .
Q5 · 2 marks · Short answerHarder board level
Evaluate $\int_{1}^{2} \frac{d x}{x ( x + 1 )}$ .
Hint
$\frac{1}{x} - \frac{1}{x + 1}$ .
Answer
$[lo g \frac{x}{x + 1}]_{1}^{2} = lo g \frac{2}{3} - lo g \frac{1}{2} = lo g \frac{4}{3}$ .
Q6 · 3 marks · Short answerHOTS / competency
Evaluate $\int \frac{d x}{e ^{x} - 1}$ .
Hint
Put $t = e^{x}$ , so $d x = \frac{d t}{t}$ .
Answer
$\int \frac{d t}{t ( t - 1 )} = \int (\frac{1}{t - 1} - \frac{1}{t}) d t = lo g \frac{t - 1}{t} = lo g ∣1 - e^{- x} ∣ + C$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a population $P$ (in thousands) grows according to $\frac{d P}{d t} = 0.01 P (100 - P)$ , with $t$ in years, so the time to grow from $P_{1}$ to $P_{2}$ is $\int_{P_{1}}^{P_{2}} \frac{d P}{0.01 P ( 100 - P )}$ .
(i) Write $\frac{1}{P ( 100 - P )}$ in partial fractions. (1 mark)
(ii) Find $\int \frac{d P}{0.01 P ( 100 - P )}$ . (1 mark)
(iii) Find the time to grow from 20 to 50 thousand. (1 mark)
(iv) Find the time to grow from 50 to 80 thousand, and compare. (1 mark)
Hint
Cover-up at $P = 0$ and $P = 100$ .
Answer
(i) $\frac{1}{100} (\frac{1}{P} + \frac{1}{100 - P})$ . (ii) $lo g \frac{P}{100 - P} + K$ . (iii) $lo g 1 - lo g \frac{20}{80} = lo g 4 \approx 1.39$ years. (iv) $lo g 4 - lo g 1 = lo g 4$ : the same time, because the model is symmetric about $P = 50$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What form do the partial fractions of $\frac{1}{( x - 1 ) ^{2} ( x + 3 )}$ take?Show answer
$\frac{A}{x - 1} + \frac{B}{( x - 1 ) ^{2}} + \frac{C}{x + 3}$ .
2.Integrate $\frac{1}{( x - 4 ) ^{2}}$ .Show answer
$- \frac{1}{x - 4} + C$ .
3.When must you divide before using partial fractions?Show answer
When the degree of the numerator is at least the degree of the denominator.
4.Find $A$ in $\frac{2 x + 1}{( x - 1 ) ( x + 2 )} = \frac{A}{x - 1} + \frac{B}{x + 2}$ .Show answer
$A = 1$ (cover-up at $x = 1$ : $\frac{3}{3}$ ).
5.Write $\int \frac{d x}{x ^{2} - 9}$ .Show answer
$\frac{1}{6} lo g \frac{x - 3}{x + 3} + C$ .
6.What numerator goes over $x^{2} + 4$ in a partial fraction?Show answer
A linear one, $B x + C$ .

Integration by Partial Fractions: frequently asked questions

What are partial fractions used for in integration?

They break a rational function into simpler fractions whose integrals are standard: logarithms, reciprocal powers and inverse tangents. Without them, most rational functions cannot be integrated directly.

How do I handle a repeated factor such as (x − 1)²?

Include one term for each power: A/(x − 1) and B/(x − 1)². Find B by putting x = 1 and A by comparing coefficients.

What if the denominator has a quadratic that cannot be factorised?

Use a linear numerator over it, (Bx + C)/(x² + bx + c). After finding B and C, split the integral into a log part and an inverse tangent part, completing the square if needed.

Is there a quick way to find the constants?

For distinct linear factors, use the cover-up rule: cover the factor and substitute its root into the rest. For other constants, compare the coefficient of the highest power and the constant term.

Which cases of partial fractions are in the ISC 2027 syllabus?

The CISCE 2027 syllabus covers proper fractions with distinct linear factors, a repeated linear factor and an irreducible quadratic factor, and improper fractions where the numerator's degree is at least the denominator's.