Partial Fractions — Breaking Complicated Fractions Apart

Why We Need Partial Fractions

Integrating $\dfrac{1}{x^2-1}$ directly is hard. But once we know it equals $\dfrac{1}{2}\left(\dfrac{1}{x-1} - \dfrac{1}{x+1}\right)$, integration becomes trivial — each term is a standard logarithm.

Partial fractions is the technique of decomposing a complicated rational function (ratio of polynomials) into a sum of simpler ones. It’s essential for integration, Laplace transforms, and solving linear recurrences.

This technique is part of CBSE Class 12 Maths (Chapter 7 — Integrals) and appears in every JEE exam.

When Can We Use Partial Fractions?

Partial fractions apply to proper rational functions — where the degree of the numerator is less than the degree of the denominator.

If the degree of the numerator ≥ degree of denominator, first perform polynomial long division to get a mixed form (quotient + proper fraction remainder), then decompose the remainder.

Example: $\dfrac{x^3 + 2}{x^2 - 1}$ — numerator degree (3) ≥ denominator degree (2), so divide first:

Then apply partial fractions to $\dfrac{x+2}{x^2-1}$.

The Forms of Partial Fractions

The form of the decomposition depends on the factors in the denominator.

Case 1 — Distinct Linear Factors

For each distinct linear factor $(ax + b)$ in the denominator, include one term $\dfrac{A}{ax+b}$.

Example:

Case 2 — Repeated Linear Factors

For a repeated linear factor $(ax+b)^n$, include $n$ terms:

Example:

Case 3 — Irreducible Quadratic Factor

For an irreducible quadratic factor $ax^2 + bx + c$ (cannot be factored into real linear factors — discriminant $< 0$), include a linear numerator $\dfrac{Ax + B}{ax^2+bx+c}$.

Example:

Case 4 — Repeated Irreducible Quadratic

For $(ax^2+bx+c)^n$, include $n$ terms with linear numerators.

Finding the Constants — Three Methods

Method 1 — Substitution (Fastest)

After multiplying both sides by the denominator, substitute the roots of each linear factor.

Example: Find A, B, C for $\dfrac{2x+1}{(x-1)(x+2)(x-3)} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3}$

Multiply both sides by $(x-1)(x+2)(x-3)$:

• Put $x=1$: $3 = A(3)(-2) = -6A \Rightarrow A = -1/2$
• Put $x=-2$: $-3 = B(-3)(-5) = 15B \Rightarrow B = -1/5$
• Put $x=3$: $7 = C(2)(5) = 10C \Rightarrow C = 7/10$

Method 2 — Comparing Coefficients

After multiplying, expand the RHS and equate coefficients of each power of $x$ from both sides. This gives a system of equations.

This method works even for irreducible quadratic factors (substituting $x = \pm i$ doesn’t give real values easily).

Method 3 — Cover-Up Rule (Quickest for Simple Cases)

For the term $\dfrac{A}{x-a}$ in $\dfrac{P(x)}{(x-a)Q(x)}$:

“Cover up” the $(x-a)$ factor in the denominator and substitute $x = a$.

Example: $\dfrac{1}{(x-1)(x+2)}$

• $A = \frac{1}{(x+2)}\big|_{x=1} = \frac{1}{3}$
• $B = \frac{1}{(x-1)}\big|_{x=-2} = \frac{1}{-3} = -\frac{1}{3}$

So $\dfrac{1}{(x-1)(x+2)} = \dfrac{1/3}{x-1} - \dfrac{1/3}{x+2} = \dfrac{1}{3}\left(\dfrac{1}{x-1} - \dfrac{1}{x+1+1}\right)$

Solved Examples

Example 1 — CBSE Level: Two Distinct Factors

Decompose $\dfrac{3x+5}{(x-1)(x+3)}$.

Solution:

$x=1$: $8 = 4A \Rightarrow A = 2$

$x=-3$: $-4 = -4B \Rightarrow B = 1$

Example 2 — JEE Level: Repeated Factor

Decompose $\dfrac{x}{(x-1)^2(x+2)}$.

Solution:

Multiply through:

$x=1$: $1 = B(3) \Rightarrow B = 1/3$

$x=-2$: $-2 = C(-3)^2 = 9C \Rightarrow C = -2/9$

$x=0$: $0 = A(-1)(2) + B(2) + C(1) = -2A + 2/3 - 2/9$

$-2A = -2/3 + 2/9 = -6/9 + 2/9 = -4/9$

$A = 2/9$

Example 3 — Integration Using Partial Fractions

Evaluate $\displaystyle\int \frac{dx}{x^2-5x+6}$.

Solution:

$x^2-5x+6 = (x-2)(x-3)$

$\dfrac{1}{(x-2)(x-3)} = \dfrac{A}{x-2} + \dfrac{B}{x-3}$

$x=2$: $1 = A(-1) \Rightarrow A = -1$

$x=3$: $1 = B(1) \Rightarrow B = 1$

Example 4 — Irreducible Quadratic

Decompose $\dfrac{x+1}{(x-1)(x^2+1)}$.

Solution:

Multiply through:

$x=1$: $2 = A(2) \Rightarrow A = 1$

Compare coefficient of $x^2$: $0 = A + B = 1 + B \Rightarrow B = -1$

Constant term: $1 = A - C = 1 - C \Rightarrow C = 0$

Common Mistakes to Avoid

Practice Questions

Q1. Decompose $\dfrac{1}{x(x-1)}$ into partial fractions.

Q2. Find: $\displaystyle\int \frac{2x+3}{(x+1)(x+2)} dx$

Q3. Decompose $\dfrac{3x-1}{(x+2)^2}$.

Q4. Evaluate $\displaystyle\int_0^1 \frac{x}{(x+1)(x+2)} dx$ (JEE level).

FAQs

How do I know if a quadratic is irreducible? Calculate the discriminant $b^2 - 4ac$. If it’s negative, the quadratic has no real roots and is irreducible over the reals. Example: $x^2 + 4$ has $b^2-4ac = 0 - 16 = -16 < 0$ → irreducible. $x^2-1$ has $b^2-4ac = 0-4(-1) = 4 > 0$ → factors as $(x-1)(x+1)$.

When is partial fractions used in real problems? Almost always in integration — whenever you see a rational function that needs to be integrated. Also in Laplace transforms (engineering and higher maths), difference equations, and sometimes in solving certain differential equations.

What if the numerator and denominator share a common factor? Cancel the common factor first (reduce the fraction), then apply partial fractions to the reduced form. Example: $\dfrac{x^2-1}{(x-1)(x+3)} = \dfrac{(x-1)(x+1)}{(x-1)(x+3)} = \dfrac{x+1}{x+3}$ (after cancellation, for $x \neq 1$) — no partial fractions needed.

Can partial fractions be used with complex numbers? Yes. Over the complex numbers, every polynomial factors into linear factors. But for real calculus applications, we keep irreducible quadratic factors as-is (to stay in real arithmetic).

Integration Using Partial Fractions — More Examples

Type: $\int \frac{px + q}{(x-a)(x-b)} dx$

Type: Repeated quadratic factor (JEE Advanced)

For $\frac{1}{(x^2+1)^2}$, we use the substitution $x = \tan\theta$ rather than partial fractions (since $x^2 + 1$ is irreducible and repeated). This gives:

This is a standard result worth memorising for JEE.

Connection to Laplace transforms

In engineering mathematics, partial fractions are essential for finding inverse Laplace transforms. The function $F(s) = \frac{1}{s(s+2)}$ in the Laplace domain corresponds to $\frac{1/2}{s} - \frac{1/2}{s+2}$, giving $f(t) = \frac{1}{2}(1 - e^{-2t})$ in the time domain. The technique is identical.

Additional Practice Questions

Q5. Evaluate $\int \frac{x}{(x+1)(x+2)(x+3)}\,dx$.

Q6. Decompose $\frac{x^2}{(x-1)(x^2+x+1)}$.

Exam Weightage

Quick decision tree for partial fractions

1. Is degree(numerator) $\geq$ degree(denominator)? If yes, do long division first.
2. Factor the denominator completely.
3. For each distinct linear factor $(x-a)$: write $\frac{A}{x-a}$.
4. For each repeated factor $(x-a)^n$: write $n$ terms with increasing powers.
5. For each irreducible quadratic: write $\frac{Ax+B}{\text{quadratic}}$.
6. Find constants by substitution (fastest) or comparing coefficients.
7. Verify by recombining.

Connection to other topics

Partial fractions connect to:

• Integration: the primary application in Class 12
• Laplace transforms: used in engineering to find inverse transforms
• Series expansion: each partial fraction can be expanded in a power series
• Difference equations: discrete analogues appear in computer science
• Number theory: the Chinese Remainder Theorem has a similar decomposition philosophy

Is there a faster way to find constants for many factors? The Heaviside cover-up method (putting $x$ equal to each root) is the fastest for distinct linear factors. For repeated or quadratic factors, compare coefficients or substitute multiple values and solve the resulting system.