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Integrals like ∫ (3x + 5) / ((x − 1)(x + 3)) dx have no direct formula. Partial fractions turn them into sums of simple pieces that integrate in one line each. Most errors come before any integration happens: the wrong form, a missing term, or forgetting to divide. This post covers the decision steps, NCERT's five forms, four worked examples and the traps.
Where is this in the syllabus?
It is section 7.5 of Chapter 7, Integrals, in the NCERT Class 12 mathematics textbook. The CBSE 2026-27 curriculum lists integration "by substitution, by partial fractions and by parts". The JEE (Main) 2026 syllabus and the JEE (Advanced) 2026 brochure both name partial fractions. Check the current year's documents.
What do you check before splitting?
Is the fraction proper? NCERT calls P(x)/Q(x) proper when the degree of P is less than the degree of Q. If not, divide first to get a polynomial plus a proper fraction.
Factorise the denominator into linear factors and quadratics that cannot be factorised further.
Pick the form for each factor.
What are the five forms?
NCERT's Table 7.2 gives these, with A, B and C as constants to find:
(px + q) / ((x − a)(x − b)) → A/(x − a) + B/(x − b)
(px + q) / (x − a)² → A/(x − a) + B/(x − a)²
Three distinct linear factors → A/(x − a) + B/(x − b) + C/(x − c)
(x − a)²(x − b) → A/(x − a) + B/(x − a)² + C/(x − b)
(x − a)(x² + bx + c), quadratic irreducible → A/(x − a) + (Bx + C)/(x² + bx + c)
Two patterns cover all five: a repeated factor gets one term for each power, and an irreducible quadratic gets a linear numerator.
How do you find the constants quickly?
Multiply through by the denominator, then substitute each root of a linear factor; each substitution kills all terms but one. For anything left, compare coefficients of the highest power or the constant term. William Bauldry describes a derivative-based way to get the same coefficients, which also handles repeated factors using higher derivatives; for exams, substitution plus one coefficient comparison is usually fastest.
Worked example 1: distinct linear factors
∫ (3x + 5) / ((x − 1)(x + 3)) dx
Write 3x + 5 = A(x + 3) + B(x − 1).
x = 1: 8 = 4A, so A = 2.
x = −3: −4 = −4B, so B = 1.
The integral is 2 ln|x − 1| + ln|x + 3| + C.
Worked example 2: a repeated factor
∫ (x² + 1) / ((x − 1)²(x + 2)) dx
Write x² + 1 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)².
x = 1: 2 = 3B, so B = 2/3.
x = −2: 5 = 9C, so C = 5/9.
Compare x² terms: 1 = A + C, so A = 4/9.
Integrate term by term. The squared term is a power, not a logarithm: ∫ (x − 1)⁻² dx = −1/(x − 1).
Answer: (4/9) ln|x − 1| − 2/(3(x − 1)) + (5/9) ln|x + 2| + C.
Worked example 3: an irreducible quadratic
∫ 2x / ((x + 1)(x² + 1)) dx
Write 2x = A(x² + 1) + (Bx + C)(x + 1).
x = −1: −2 = 2A, so A = −1.
Compare x² terms: 0 = A + B, so B = 1.
Compare constants: 0 = A + C, so C = 1.
So the integrand is −1/(x + 1) + (x + 1)/(x² + 1). Split the last piece into x/(x² + 1) + 1/(x² + 1).
Answer: −ln|x + 1| + ½ ln(x² + 1) + tan⁻¹ x + C.
Worked example 4: an improper fraction
∫ (x² + x + 1) / (x² − 1) dx
The degrees are equal, so divide: (x² + x + 1) / (x² − 1) = 1 + (x + 2) / (x² − 1).
Then (x + 2) / ((x − 1)(x + 1)) = A/(x − 1) + B/(x + 1), with x = 1 giving 3 = 2A (A = 3/2) and x = −1 giving 1 = −2B (B = −1/2).
Answer: x + (3/2) ln|x − 1| − (1/2) ln|x + 1| + C.
Which mistakes cost marks?
Skipping the division for an improper fraction. The constants will not work out.
Missing a power of a repeated factor: (x − a)² needs both A/(x − a) and B/(x − a)².
A constant over a quadratic instead of Bx + C.
Integrating 1/(x − a)² as a logarithm. It is −1/(x − a).
Dropping the absolute value in ln|x − a|.
Not checking. Put a value such as x = 0 into both the original and your split; they must agree.
A practice routine
Write NCERT's five forms from memory until they are automatic.
Do NCERT Exercise 7.5 in mixed order rather than straight through.
Shuffle partial fraction problems in with substitution and by-parts integrals. In Rohrer and Taylor's experiments, students whose practice was mixed did far better on a test a week later than students who practised one type at a time. Our integration practice plan and the post on choosing the first function in integration by parts fit into the same mix.
If you do not know where to begin with integration as a whole, Fermi's AI tutor works out what you already know and builds a plan for the session, so you start with the cases you find hardest.
Tonight, write the partial fraction form, without constants, for five denominators of your own, one of each type in Table 7.2.
