Integration by Partial Fractions: Class 12 Maths Step-by-Step Guide

Integration by Partial Fractions: Class 12 Maths Step-by-Step Guide

When the integrand is a rational function — a ratio of two polynomials — and none of the standard formulas or trigonometric identities apply, partial fractions is the technique to reach for. The core idea is elegant: instead of integrating one complicated fraction, you break it down into a sum of simpler fractions, each of which integrates using formulas you already know. This guide covers every standard case tested in CBSE Class 12 boards, with the exact algebraic steps to avoid the calculation traps that cost students marks.

What Is a Rational Function, and Why Does It Need Special Treatment?

A rational function has the form P(x)/Q(x), where both P(x) and Q(x) are polynomials and Q(x) ≠ 0. If the degree of P(x) is less than the degree of Q(x), the function is called proper; otherwise, it's improper. Improper rational functions must first be reduced using long division: P(x)/Q(x) = T(x) + P₁(x)/Q(x), where T(x) is a polynomial (easy to integrate directly) and P₁(x)/Q(x) is now a proper fraction. Only proper rational functions require partial fraction decomposition.

The Five Standard Cases

The exact form of the partial fraction decomposition depends entirely on how the denominator Q(x) factors:

Correctly identifying which case applies — based on whether factors are linear, repeated, or an irreducible quadratic — is the first and most important step.

Case 1: Distinct Linear Factors

Example 1: Evaluate ∫dx/[(x+1)(x+2)]

Write 1/[(x+1)(x+2)] = A/(x+1) + B/(x+2). Multiplying through: 1 = A(x+2) + B(x+1).

Comparing coefficients of x and the constant term: A + B = 0 and 2A + B = 1. Solving gives A = 1, B = −1.

So the integral becomes: ∫dx/(x+1) − ∫dx/(x+2) = log|x+1| − log|x+2| + C = log|(x+1)/(x+2)| + C

Faster shortcut: For distinct linear factors, you can also find A and B by substituting convenient values of x directly into 1 = A(x+2) + B(x+1) — for instance, x = −1 gives A = 1 immediately, and x = −2 gives B = −1. This "cover-up" trick saves time in exams.

Case 2: Repeated Linear Factors

Example 2: Evaluate ∫(3x−2)/(x+1)² dx

Write (3x−2)/(x+1)² = A/(x+1) + B/(x+1)². Then 3x − 2 = A(x+1) + B.

Comparing coefficients of x: A = 3. Substituting x = −1: 3(−1) − 2 = B, so B = −5.

= 3∫dx/(x+1) − 5∫dx/(x+1)² = 3 log|x+1| + 5/(x+1) + C

Notice the sign flip: since ∫(x+1)⁻² dx = −(x+1)⁻¹ + C, multiplying by −5 gives +5/(x+1).

Case 3: Three Distinct Linear Factors

Example 3: Evaluate ∫dx/[(x−1)(x+2)(x−3)]

Set up: 1/[(x−1)(x+2)(x−3)] = A/(x−1) + B/(x+2) + C/(x−3)

Using the cover-up method: A = 1/[(1+2)(1−3)] = 1/(3×−2) = −1/6

B = 1/[(−2−1)(−2−3)] = 1/((−3)(−5)) = 1/15

C = 1/[(3−1)(3+2)] = 1/(2×5) = 1/10

So the integral is: −(1/6)log|x−1| + (1/15)log|x+2| + (1/10)log|x−3| + C

This case involves the most arithmetic, so double-checking each substitution carefully is essential to avoid sign errors.

Case 5: Irreducible Quadratic Factors

When the denominator contains a quadratic that cannot be factored into real linear terms (its discriminant is negative), the numerator over that factor must be linear (Bx + C), not just a constant.

Example 4: Evaluate ∫(x²+x+1)/[(x+2)(x²+1)] dx

Write x² + x + 1 = A(x²+1) + (Bx+C)(x+2). Expanding and comparing coefficients of x², x, and the constant term gives three equations: A + B = 1, 2B + C = 1, A + 2C = 1. Solving simultaneously: A = 3/5, B = 2/5, C = 1/5.

The integral splits into: (3/5)∫dx/(x+2) + (1/5)∫(2x+1)/(x²+1) dx

= (3/5)log|x+2| + (1/5)log(x²+1) + (1/5)tan⁻¹x + C

Notice how the (Bx+C)/(x²+1) piece further splits into a logarithmic part (from the 2x term, matching the derivative of the denominator) and an inverse tangent part (from the constant term).

A Trick for Even-Power Denominators

Some problems, like ∫x² dx/[(x²+1)(x²+4)], become much easier if you first substitute y = x². This turns the rational function into y/[(y+1)(y+4)], which is a simple Case 1 problem in y. After decomposing and integrating with respect to y, substitute back y = x² and use the standard tan⁻¹ formula. This shortcut avoids unnecessarily messy quartic algebra.

Suggested Internal Links

• Link to: Standard Integral Forms: x²±a² Formulas (used for the tan⁻¹ portion of quadratic-factor cases)
• Link to: Integration by Substitution Method (for combined substitution + partial fraction problems)
• Link to: Integration by Parts – Complete Guide (the next major technique in this chapter)
• Link to: CBSE Class 12 Maths Question Bank (for extra practice sets)

Ready to Practice Partial Fractions?

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