MathsCBSE Class 12 · Integrals

Integration by Partial Fractions: All Three Forms with Solved Questions (Class 12)

Q: When should I use partial fractions?

When the integrand is a rational function (a polynomial divided by a polynomial) whose denominator can be factorised, and no simpler substitution works.

Q: How do I find the constants quickly?

Multiply through by the denominator, then put x equal to each root of a linear factor; this gives those constants immediately. Compare coefficients (for example of x² or the constant term) for any constants that remain.

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

Use a numerator of the form Bx + C over that quadratic. After finding B and C, split it into a log part and an inverse tangent part.

Q: Which NCERT section covers partial fractions?

Section 7.5 (Integration by Partial Fractions) of Chapter 7, Integrals, in NCERT Mathematics Part II for Class 12, with Table 7.2 listing the forms.

Q: Is long division needed often?

Whenever the degree of the numerator is equal to or greater than the degree of the denominator. Skipping it is a common reason for losing all the marks.

Updated 15 October 2026 · Fact-checked

To integrate a rational function P(x)/Q(x), first make it proper (degree of P less than degree of Q) by long division. Then factorise Q and split 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.

Factorise the denominator first, then pick the matching form: distinct linear, repeated linear, or linear with an unfactorable quadratic.

Illustration drawn for QuestionBuddy.

Understand Integration by Partial Fractions

A rational function is a quotient of polynomials, $\frac{P ( x )}{Q ( x )}$ . Most cannot be integrated as they stand, but they can be split into simple fractions whose integrals are logs, powers or inverse tangents.

Step 0: proper or improper? If $de g P \geq de g Q$ , the fraction is improper; divide first, writing $\frac{P}{Q} = (polynomial) + \frac{R}{Q}$ with $de g R < de g Q$ .

Step 1: factorise the denominator. Then use the matching form (NCERT Table 7.2):

Factor in denominator	Partial fraction terms
$(x - a) (x - b)$ , $a \neq = b$	$\frac{A}{x - a} + \frac{B}{x - b}$
$(x - a)^{2}$	$\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}}$
$(x - a) (x^{2} + b x + c)$ , quadratic not factorable	$\frac{A}{x - a} + \frac{B x + C}{x ^{2} + b x + c}$

Step 2: find the constants. Multiply through by the denominator to get a polynomial identity. Then either put convenient values of $x$ (the roots make most terms vanish) or compare coefficients of equal powers; combine both for repeated or quadratic factors.

Step 3: integrate. $\int \frac{d x}{x - a} = lo g ∣ x - a ∣$ , $\int \frac{d x}{( x - a ) ^{2}} = - \frac{1}{x - a}$ , and $\frac{B x + C}{x ^{2} + k ^{2}}$ splits into $\frac{B}{2} \cdot \frac{2 x}{x ^{2} + k ^{2}}$ (a log) plus $\frac{C}{x ^{2} + k ^{2}}$ (an inverse tangent). Combine logs at the end if you like: $lo g ∣ x - 1∣ - lo g ∣ x + 1∣ = lo g \frac{x - 1}{x + 1}$ .

Substitution first. Many integrals become rational only after a substitution: $\int \frac{e ^{x} d x}{( e ^{x} + 1 ) ( e ^{x} + 3 )}$ with $t = e^{x}$ , or $\int \frac{d x}{x ( x ^{4} + 1 )}$ , which becomes $\frac{1}{4} \int \frac{d t}{t ( t + 1 )}$ after multiplying top and bottom by $x^{3}$ and putting $t = x^{4}$ . Trigonometric versions, such as $\frac{c o s x}{( 1 - s i n x ) ( 2 - s i n x )}$ , work the same way with $t = sin x$ .

A shortcut for two linear factors. For $\frac{1}{( x - a ) ( x - b )}$ , $A = \frac{1}{a - b}$ and $B = \frac{1}{b - a}$ (the 'cover-up' rule: cover a factor and put its root into the rest).

Partial fractions is NCERT Section 7.5 and appeared in the 2023, 2024, 2025 and 2026 board papers as 3- and 5-mark questions.

Integration by Partial Fractions: formulas to remember

Distinct linear factors: $\frac{p x + q}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}$; Find A and B by putting $x = a$ and $x = b$ .
Repeated linear factor: $\frac{p x ^{2} + q x + r}{( x - a ) ^{2} ( x - b )} = \frac{A}{x - a} + \frac{B}{( x - a ) ^{2}} + \frac{C}{x - b}$; Both powers of the repeated factor appear.
Irreducible quadratic factor: $\frac{p x ^{2} + q x + r}{( x - a ) ( x ^{2} + b x + c )} = \frac{A}{x - a} + \frac{B x + C}{x ^{2} + b x + c}$; A linear numerator over the quadratic.
Basic integrals: $\int \frac{d x}{x - a} = lo g ∣ x - a ∣, \int \frac{d x}{( x - a ) ^{2}} = - \frac{1}{x - a}$; Add $+ C$ at the end.
Quadratic pieces: $\int \frac{x d x}{x ^{2} + k ^{2}} = \frac{1}{2} lo g (x^{2} + k^{2}), \int \frac{d x}{x ^{2} + k ^{2}} = \frac{1}{k} tan^{- 1} \frac{x}{k}$; Split $\frac{B x + C}{x ^{2} + k ^{2}}$ into these two.
Standard results: $\int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} lo g \frac{x - a}{x + a} + C$; A partial-fraction result worth memorising.

How to solve Integration by Partial Fractions questions

A dependable five-step routine for any rational integrand.

1If the degree of the numerator is not less than that of the denominator, divide first.
2Factorise the denominator completely into linear and irreducible quadratic factors.
3Write the partial-fraction form with unknown constants, using the table of forms.
4Multiply by the denominator and find the constants by substituting roots and comparing coefficients.
5Integrate term by term, combine logs if neat, and add $+ C$ (or apply limits for a definite integral).

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.

Total amount from a rational rate

Suppose water flows into a small tank at the rate $r (t) = \frac{12}{( t + 1 ) ( t + 3 )}$ litres per minute, $t$ minutes after a tap is opened.

How much water flows in during the first minute?

Try it first, then show the solutionSolution

$\frac{12}{( t + 1 ) ( t + 3 )} = \frac{6}{t + 1} - \frac{6}{t + 3}$ (cover-up: $t = - 1$ gives 6, $t = - 3$ gives $- 6$ ).
$\int_{0}^{1} r (t) d t = 6 [lo g (t + 1) - lo g (t + 3)]_{0}^{1} = 6 [lo g \frac{2}{4} - lo g \frac{1}{3}]$ .
$= 6 lo g \frac{3}{2} \approx 2.43$ litres.

Answer: $6 lo g \frac{3}{2} \approx 2.43$ litres

Memory hook: Split the rate into simple fractions, then each piece integrates to a log.

Distance from a slowing speed

Imagine a toy car whose speed is $v (t) = \frac{20}{t ^{2} + 5 t + 6}$ metres per second after its motor is switched off.

Find the distance it covers in the first 2 seconds.

Try it first, then show the solutionSolution

$t^{2} + 5 t + 6 = (t + 2) (t + 3)$ , so $v = \frac{20}{t + 2} - \frac{20}{t + 3}$ .
Distance $= 20 [lo g (t + 2) - lo g (t + 3)]_{0}^{2} = 20 [lo g \frac{4}{5} - lo g \frac{2}{3}]$ .
$= 20 lo g \frac{6}{5} \approx 3.65$ m.

Answer: $20 lo g \frac{6}{5} \approx 3.65$ m

Memory hook: Factor the quadratic, split, integrate: speed becomes distance.

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 only work on proper fractions.
Use it when: Improper rational functions.
Cover-up for linear factors
To find the constant over $(x - a)$ , cover that factor and put $x = a$ in what is left.
Use it when: Fast constants for distinct linear factors.
Repeated factor, every power
$(x - a)^{2}$ needs both $\frac{A}{x - a}$ and $\frac{B}{( x - a ) ^{2}}$ .
Use it when: Repeated-factor questions.
Quadratic gets a linear top
Over an irreducible quadratic, the numerator is $B x + C$ , not just a constant.
Use it when: Denominators like $(x + 1) (x^{2} + 4)$ .

Integration by Partial Fractions: solved examples

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

Example 1: Two distinct linear factors · 3 marks · Short answerBoard level

Evaluate $\int \frac{3 x + 5}{( x - 1 ) ( x + 3 )} d x$ .

Try it first, then show the solutionSolution

$\frac{3 x + 5}{( x - 1 ) ( x + 3 )} = \frac{A}{x - 1} + \frac{B}{x + 3}$ , so $3 x + 5 = A (x + 3) + B (x - 1)$ . (1 mark)
$x = 1$ : $8 = 4 A \Rightarrow A = 2$ . $x = - 3$ : $- 4 = - 4 B \Rightarrow B = 1$ . (1 mark)
$\int (\frac{2}{x - 1} + \frac{1}{x + 3}) d x = 2 lo g ∣ x - 1∣ + lo g ∣ x + 3∣ + C$ . (1 mark)

Answer: $2 lo g ∣ x - 1∣ + lo g ∣ x + 3∣ + C$

Example 2: A repeated linear factor · 3 marks · Short answerBoard level

Evaluate $\int \frac{x + 3}{( x + 1 ) ^{2} ( x - 1 )} d x$ .

Try it first, then show the solutionSolution

$\frac{x + 3}{( x + 1 ) ^{2} ( x - 1 )} = \frac{A}{x + 1} + \frac{B}{( x + 1 ) ^{2}} + \frac{C}{x - 1}$ , so $x + 3 = A (x + 1) (x - 1) + B (x - 1) + C (x + 1)^{2}$ . (1 mark)
$x = 1$ : $4 = 4 C \Rightarrow C = 1$ . $x = - 1$ : $2 = - 2 B \Rightarrow B = - 1$ . Coefficient of $x^{2}$ : $0 = A + C \Rightarrow A = - 1$ . (1 mark)
$\int (\frac{- 1}{x + 1} - \frac{1}{( x + 1 ) ^{2}} + \frac{1}{x - 1}) d x = - lo g ∣ x + 1∣ + \frac{1}{x + 1} + lo g ∣ x - 1∣ + C$ . (1 mark)

Answer: $lo g \frac{x - 1}{x + 1} + \frac{1}{x + 1} + C$

Example 3: A linear and an irreducible quadratic factor · 5 marks · Long answerHarder board level

Evaluate $\int \frac{5 x}{( x + 1 ) ( x ^{2} + 4 )} d x$ .

Try it first, then show the solutionSolution

$\frac{5 x}{( x + 1 ) ( x ^{2} + 4 )} = \frac{A}{x + 1} + \frac{B x + C}{x ^{2} + 4}$ , so $5 x = A (x^{2} + 4) + (B x + C) (x + 1)$ . (1 mark)
$x = - 1$ : $- 5 = 5 A \Rightarrow A = - 1$ . Coefficient of $x^{2}$ : $0 = A + B \Rightarrow B = 1$ . Constant term: $0 = 4 A + C \Rightarrow C = 4$ . (1½ marks)
Integrand $= - \frac{1}{x + 1} + \frac{x}{x ^{2} + 4} + \frac{4}{x ^{2} + 4}$ . (½ mark)
$\int = - lo g ∣ x + 1∣ + \frac{1}{2} lo g (x^{2} + 4) + 4 \cdot \frac{1}{2} tan^{- 1} \frac{x}{2} + C$ . (1½ marks)
$= - lo g ∣ x + 1∣ + \frac{1}{2} lo g (x^{2} + 4) + 2 tan^{- 1} \frac{x}{2} + C$ . (½ mark)

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

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

Evaluate $\int \frac{x ^{3} + 2}{x ^{2} - 4} d x$ .

Try it first, then show the solutionSolution

Divide: $x^{3} + 2 = x (x^{2} - 4) + 4 x + 2$ , so the integrand is $x + \frac{4 x + 2}{( x - 2 ) ( x + 2 )}$ . (1 mark)
$\frac{4 x + 2}{( x - 2 ) ( x + 2 )} = \frac{A}{x - 2} + \frac{B}{x + 2}$ : $x = 2$ gives $A = \frac{10}{4} = \frac{5}{2}$ ; $x = - 2$ gives $B = \frac{- 6}{- 4} = \frac{3}{2}$ . (1 mark)
$\int = \frac{x ^{2}}{2} + \frac{5}{2} lo g ∣ x - 2∣ + \frac{3}{2} lo g ∣ x + 2∣ + C$ . (1 mark)

Answer: $\frac{x ^{2}}{2} + \frac{5}{2} lo g ∣ x - 2∣ + \frac{3}{2} lo g ∣ x + 2∣ + C$

Example 5: Substitute, then split · 3 marks · Short answerHOTS / competency

Evaluate $\int \frac{e ^{x}}{( e ^{x} + 1 ) ( e ^{x} + 3 )} d x$ .

Try it first, then show the solutionSolution

Put $t = e^{x}$ , $d t = e^{x} d x$ : $\int \frac{d t}{( t + 1 ) ( t + 3 )}$ . (1 mark)
$\frac{1}{( t + 1 ) ( t + 3 )} = \frac{1}{2} (\frac{1}{t + 1} - \frac{1}{t + 3})$ . (1 mark)
$\int = \frac{1}{2} lo g \frac{t + 1}{t + 3} + C = \frac{1}{2} lo g (\frac{e ^{x} + 1}{e ^{x} + 3}) + C$ . (1 mark)

Answer: $\frac{1}{2} lo g (\frac{e ^{x} + 1}{e ^{x} + 3}) + C$

Example 6: Case study: a medicine dose over time · 4 marks · Case-basedHOTS / competency

Suppose a slow-release tablet delivers medicine at the rate $r (t) = \frac{10}{( t + 1 ) ( t + 2 )}$ mg per hour, $t$ hours after it is swallowed.
(i) Write $r (t)$ in partial fractions. (1 mark)
(ii) Find $\int r (t) d t$ . (1 mark)
(iii) How much medicine is delivered in the first 3 hours? (1 mark)
(iv) What total does the delivery approach in the long run? (1 mark)

Try it first, then show the solutionSolution

(i) $\frac{10}{( t + 1 ) ( t + 2 )} = \frac{10}{t + 1} - \frac{10}{t + 2}$ . (1 mark)
(ii) $\int r (t) d t = 10 lo g \frac{t + 1}{t + 2} + C$ . (1 mark)
(iii) $10 [lo g \frac{4}{5} - lo g \frac{1}{2}] = 10 lo g \frac{8}{5} \approx 4.70$ mg. (1 mark)
(iv) Over $[0, T]$ the total is $10 [lo g \frac{T + 1}{T + 2} + lo g 2]$ ; as $T$ grows, $\frac{T + 1}{T + 2} \to 1$ , so the total approaches $10 lo g 2 \approx 6.93$ mg. (1 mark)

Answer: (i) $\frac{10}{t + 1} - \frac{10}{t + 2}$ (ii) $10 lo g \frac{t + 1}{t + 2} + C$ (iii) $10 lo g \frac{8}{5} \approx 4.70$ mg (iv) $10 lo g 2 \approx 6.93$ mg.

Where students lose marks in Integration by Partial Fractions

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

Mistake: Splitting an improper fraction without dividing
Partial fractions apply only to proper fractions.
Wrong working
$\frac{x ^{3} + 2}{x ^{2} - 4} = \frac{A}{x - 2} + \frac{B}{x + 2}$ .
Right working
$\frac{x ^{3} + 2}{x ^{2} - 4} = x + \frac{4 x + 2}{x ^{2} - 4}$ , then split the proper part.
Do this: Compare degrees before anything else.
Usually costs: Most of the marks.
Mistake: Missing a term for a repeated factor
$(x + 1)^{2}$ needs both $\frac{A}{x + 1}$ and $\frac{B}{( x + 1 ) ^{2}}$ .
Wrong working
$\frac{x + 3}{( x + 1 ) ^{2} ( x - 1 )} = \frac{B}{( x + 1 ) ^{2}} + \frac{C}{x - 1}$ .
Right working
Include $\frac{A}{x + 1}$ ; here $A = - 1$ .
Do this: List one term for every power of a repeated factor.
Usually costs: 2 to 3 marks.
Mistake: Constant numerator over a quadratic
An irreducible quadratic needs $B x + C$ on top.
Wrong working
$\frac{5 x}{( x + 1 ) ( x ^{2} + 4 )} = \frac{A}{x + 1} + \frac{C}{x ^{2} + 4}$ .
Right working
$\frac{A}{x + 1} + \frac{B x + C}{x ^{2} + 4}$ with $A = - 1$ , $B = 1$ , $C = 4$ .
Do this: Linear on top of every irreducible quadratic.
Usually costs: The constants and the final answer.
Mistake: Wrong sign integrating 1/(x − a)²
$\int (x - a)^{- 2} d x = - (x - a)^{- 1}$ .
Wrong working
$\int \frac{- 1}{( x + 1 ) ^{2}} d x = - \frac{1}{x + 1}$ .
Right working
$\int \frac{- 1}{( x + 1 ) ^{2}} d x = \frac{1}{x + 1}$ .
Do this: Use the power rule carefully, or differentiate your answer to check.
Usually costs: Half to one mark.

Exam tips

Write the partial-fraction form with unknowns before solving; the form itself earns a mark.
Use substitution of roots first, then compare coefficients for the remaining constants.
Check constants by putting a simple value such as $x = 0$ into both sides.
Present the final answer as separate log terms or one combined log; both are accepted.
Look out for disguised rational functions in $e^{x}$ , $sin x$ or $x^{n}$ and substitute first.

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
$\int \frac{d x}{x ^{2} - 9}$ equals:
(a) $\frac{1}{3} tan^{- 1} \frac{x}{3} + C$
(b) $\frac{1}{6} lo g \frac{x - 3}{x + 3} + C$
(c) $\frac{1}{6} lo g \frac{x + 3}{x - 3} + C$
(d) $lo g ∣ x^{2} - 9∣ + C$
Hint
$\frac{1}{x ^{2} - a ^{2}} = \frac{1}{2 a} (\frac{1}{x - a} - \frac{1}{x + a})$ .
Answer
(b).
Q2 · 2 marks · Very short answerBoard level
Evaluate $\int \frac{d x}{x ( x + 1 )}$ .
Hint
$\frac{1}{x ( x + 1 )} = \frac{1}{x} - \frac{1}{x + 1}$ .
Answer
$lo g ∣ x ∣ - lo g ∣ x + 1∣ + C = lo g \frac{x}{x + 1} + C$ .
Q3 · 3 marks · Short answerBoard level
Evaluate $\int \frac{x}{( x - 2 ) ( x - 3 )} d x$ .
Hint
Use $x = 2$ and $x = 3$ in $x = A (x - 3) + B (x - 2)$ .
Answer
$A = - 2$ , $B = 3$ : $- 2 lo g ∣ x - 2∣ + 3 lo g ∣ x - 3∣ + C$ .
Q4 · 3 marks · Short answerHarder board level
Evaluate $\int \frac{2 x + 1}{( x + 2 ) ( x - 3 ) ^{2}} d x$ .
Hint
Form: $\frac{A}{x + 2} + \frac{B}{x - 3} + \frac{C}{( x - 3 ) ^{2}}$ .
Answer
$2 x + 1 = A (x - 3)^{2} + B (x + 2) (x - 3) + C (x + 2)$ . $x = 3$ : $C = \frac{7}{5}$ ; $x = - 2$ : $A = - \frac{3}{25}$ ; $x^{2}$ : $B = \frac{3}{25}$ . Integral $= - \frac{3}{25} lo g ∣ x + 2∣ + \frac{3}{25} lo g ∣ x - 3∣ - \frac{7}{5 ( x - 3 )} + C$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{x ^{2}}{( x - 1 ) ( x - 2 )}$ cannot be written directly as $\frac{A}{x - 1} + \frac{B}{x - 2}$ .
Reason (R): The degree of the numerator is not less than the degree of the denominator.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What does $\frac{A}{x - 1} + \frac{B}{x - 2}$ give as $x \to \infty$ ?
Answer
(a). The fraction is improper, so a polynomial part (here 1) must be separated by division first: $\frac{x ^{2}}{( x - 1 ) ( x - 2 )} = 1 + \frac{3 x - 2}{( x - 1 ) ( x - 2 )}$ .
Q6 · 2 marks · Very short answerHarder board level
Evaluate $\int_{1}^{2} \frac{d x}{x ( x + 1 )}$ .
Hint
Use the antiderivative $lo g \frac{x}{x + 1}$ .
Answer
$[lo g \frac{x}{x + 1}]_{1}^{2} = lo g \frac{2}{3} - lo g \frac{1}{2} = lo g \frac{4}{3}$ .
Q7 · 3 marks · Short answerHOTS / competency
Evaluate $\int \frac{d x}{x ( x ^{4} + 1 )}$ .
Hint
Multiply top and bottom by $x^{3}$ , then put $t = x^{4}$ .
Answer
$\int \frac{x ^{3} d x}{x ^{4} ( x ^{4} + 1 )}$ ; $t = x^{4}$ , $d t = 4 x^{3} d x$ : $\frac{1}{4} \int \frac{d t}{t ( t + 1 )} = \frac{1}{4} lo g \frac{t}{t + 1} + C = \frac{1}{4} lo g \frac{x ^{4}}{x ^{4} + 1} + C$ .

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 is the first check before using partial fractions?Show answer
Whether the fraction is proper; if not, divide.
2.Partial-fraction form for $\frac{1}{( x - a ) ^{2} ( x - b )}$ ?Show answer
$\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}} + \frac{C}{x - b}$ .
3.Form for $\frac{1}{( x - a ) ( x ^{2} + 1 )}$ ?Show answer
$\frac{A}{x - a} + \frac{B x + C}{x ^{2} + 1}$ .
4.What is $\int \frac{d x}{( x - 2 ) ^{2}}$ ?Show answer
$- \frac{1}{x - 2} + C$ .
5.Split $\frac{1}{x ( x - 1 )}$ .Show answer
$\frac{1}{x - 1} - \frac{1}{x}$ .
6.What is $\int \frac{d x}{x ^{2} - a ^{2}}$ ?Show answer
$\frac{1}{2 a} lo g \frac{x - a}{x + a} + C$ .

Integration by Partial Fractions: frequently asked questions

When should I use partial fractions?

When the integrand is a rational function (a polynomial divided by a polynomial) whose denominator can be factorised, and no simpler substitution works.

How do I find the constants quickly?

Multiply through by the denominator, then put x equal to each root of a linear factor; this gives those constants immediately. Compare coefficients (for example of x² or the constant term) for any constants that remain.

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

Use a numerator of the form Bx + C over that quadratic. After finding B and C, split it into a log part and an inverse tangent part.

Which NCERT section covers partial fractions?

Section 7.5 (Integration by Partial Fractions) of Chapter 7, Integrals, in NCERT Mathematics Part II for Class 12, with Table 7.2 listing the forms.

Is long division needed often?

Whenever the degree of the numerator is equal to or greater than the degree of the denominator. Skipping it is a common reason for losing all the marks.