Adding x−12+x+23 over a common denominator gives x2+x−25x+1. Partial fraction decomposition reverses this — it breaks one complicated rational function back into a sum of simple pieces. Those pieces are far easier to integrate: ∫x−12dx=2log∣x−1∣ is immediate, while the combined fraction is not.
When it applies
You need a proper rational function, degP<degQ. If the numerator's degree is equal or higher, first do polynomial long division. The denominator Q(x) must be factored into linear and/or irreducible quadratic factors.
The standard forms
For Q(x)P(x) with Q fully factored, each factor contributes a term:
Distinct linear (ax+b)→ax+bA.
Repeated linear (ax+b)n→ax+bA1+(ax+b)2A2+⋯+(ax+b)nAn.
Irreducible quadratic (ax2+bx+c)→ax2+bx+cAx+B — a linear numerator, not just a constant.
The unknown constants are found from the resulting equations; the decomposition is unique, which is what lets us solve for them systematically.
Worked example
Decompose x2−3x+23x+5. Factor the denominator: (x−1)(x−2). Set
(x−1)(x−2)3x+5=x−1A+x−2B.
Clear denominators: 3x+5=A(x−2)+B(x−1). Substituting the roots, x=1 gives 8=−A so A=−8, and x=2 gives 11=B. Hence
x2−3x+23x+5=x−1−8+x−211.
Tip
With distinct linear factors, substitute each factor's root to knock out all but one term — much faster than equating coefficients.
Watch out
Match the numerator to the factor: a quadratic factor needs Ax+B, and a repeated factor needs a term for every power up to its multiplicity.
In Class 12 the main use is integration — every rational function can be integrated once decomposed this way.
Partial fraction decomposition is its own dedicated section in the NCERT Class 12 Integrals chapter, and it's one of the most frequently tested multi-step problems in CBSE boards and JEE Main integration questions. Students searching 'partial fractions integration class 12 examples' or 'partial fraction decomposition formula for repeated and quadratic factors' will find this break-into-simple-terms method is exactly the standard procedure those exam solutions follow.
Concept: Partial Fraction Decomposition — splitting a rational function into simpler fractions that integrate term-by-term to logs.
We want
∫(x−1)(x−2)(x−3)3x−1dx
Step 1 – Set up the decomposition
Since the denominator has three distinct linear factors, write
We decompose the rational function into partial fractions of the form x−1A+x−2B+x−3C, solve for A,B,C using the cover-up method, then integrate each term to get log(x−2)(x−3)(x−1)2+C.
Why Partial Fractions?
When you have a rational function where the denominator factors into distinct linear factors, integration becomes straightforward if you can split it into a sum of simpler fractions. Each term x−aA integrates to Alog∣x−a∣, which is clean and easy. The trick is finding the right constants A,B,C so that the sum equals the original fraction.
The denominator here is (x−1)(x−2)(x−3) — three distinct linear factors. That means we can write:
(x−1)(x−2)(x−3)3x−1=x−1A+x−2B+x−3C
where A,B,C are constants to be determined.
Step-by-step solution
1. Set up the equation
Multiply both sides by the denominator (x−1)(x−2)(x−3) to clear fractions:
3x−1=A(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)
This identity must hold for all x.
2. Use the cover-up method for each constant
Since the factors are linear and distinct, we can find each constant by substituting the root that makes the other terms vanish.
For A: Set x=1. Then (x−1)=0, so the B and C terms disappear.
3(1)−1=A(1−2)(1−3)⟹2=A(−1)(−2)=2A⟹A=1
For B: Set x=2.
3(2)−1=B(2−1)(2−3)⟹5=B(1)(−1)=−B⟹B=−5
For C: Set x=3.
3(3)−1=C(3−1)(3−2)⟹8=C(2)(1)=2C⟹C=4
Tip
The cover-up method works because when you plug x=a, all terms except the one with (x−a) in the denominator vanish — the factor (x−a) multiplies the other terms to zero. It's the fastest way for distinct linear factors.
3. Write the partial fraction decomposition
We now have:
(x−1)(x−2)(x−3)3x−1=x−11−x−25+x−34
Watch out
A common mistake is forgetting the sign when B comes out negative. Double-check: B=−5 means the term is −x−25, not +x−25.
This is a perfectly acceptable final form. Some textbooks prefer to keep it as a sum of logs, but the compact single-log form is cleaner.
Note
The absolute values are necessary because the domain of the original function excludes x=1,2,3, and the logarithm is only defined for positive arguments. The absolute value ensures the expression is valid on each interval of the domain.
✓Final answer
The integral is log(x−2)5(x−1)(x−3)4+C, or equivalently log∣x−1∣−5log∣x−2∣+4log∣x−3∣+C.
Method: Partial fractions with three distinct linear factors
Use this for (x−a)(x−b)(x−c)N(x) with a numerator of lower degree — three different linear factors, three constants, three logarithms.
Steps
Step 1: One constant per factor.
(x−a)(x−b)(x−c)N(x)=x−aA+x−bB+x−cC.
Step 2: Clear denominators to get N(x)=A(x−b)(x−c)+B(x−a)(x−c)+C(x−a)(x−b).
Step 3: Cover-up at each root. Substituting x=a kills the B and C terms, giving A immediately; likewise x=b gives B and x=c gives C.
Carry each sign exactly and keep the absolute values.
Common Mistakes
Mistake 1: Sign error when a constant comes out negative.
Why it's wrong: here B=−5, so the middle term is −5log∣x−2∣; writing +5 flips it. Correct approach: at x=2, 5=B(1)(−1)=−B, so B=−5 — carry the sign through to the integral.
Mistake 2: Multiplying the wrong bracket values in cover-up.
Why it's wrong: at x=1, A=(x−2)(x−3)3x−1=(−1)(−2)2=1; using (1−2)(1−3) with a sign slip gives A=−1. Correct approach: substitute the root into every remaining factor, minding each sign.
Mistake 3: Dropping absolute values or the constant of integration.
Why it's wrong: ∫x−adx=log∣x−a∣, defined only away from x=1,2,3; missing bars or C leaves the antiderivative incomplete. Correct approach: write log∣⋅∣ for each term and add +C.