Integration by partial fractions

Some fractions refuse to integrate directly no matter how hard you stare at them — $\dfrac{3x+5}{(x-1)(x+2)}$ is not the derivative of anything obvious, and no substitution simplifies it. Integration by partial fractions is the fix: split a rational function $\dfrac{P(x)}{Q(x)}$ into a sum of simpler fractions — one per factor of the denominator — each of which integrates to a standard logarithm or an inverse-tangent form. This is one of the three core techniques in the Integrals chapter of CBSE Class 12 Mathematics, used whenever the integrand is a ratio of polynomials with no clean derivative-of-denominator relationship. The syllabus keeps to three factor patterns: distinct linear factors, repeated linear factors, and irreducible quadratic factors. Master the decomposition templates and two ways to pin down the unknown constants — substitution and coefficient comparison — and the integral itself becomes routine.

The core insight

Split $\dfrac{P(x)}{Q(x)}$ into a sum of simple fractions matching the factors of $Q(x)$ , then integrate each piece using the standard log and arctan forms.

The core idea

Proper vs improper. A rational function $\dfrac{P(x)}{Q(x)}$ is proper when the degree of $P$ is less than the degree of $Q$ . If it is improper (degree of $P \ge$ degree of $Q$ ), do polynomial long division first, writing it as a quotient polynomial plus a proper remainder fraction, and apply partial fractions only to that remainder.

Factor and template. Factor $Q(x)$ completely, then assign one partial-fraction term per factor, with constants $A, B, \dots$ to be found:

\frac{P(x)}{Q(x)} = \frac{A}{x-a} \;+\; \frac{B}{x-b} \;+\; \frac{Cx+D}{ex^2+fx+g} \;+\; \cdots

where: each linear factor $(x-a)$ contributes a single term $\dfrac{A}{x-a}$ ; a repeated linear factor $(x-a)^2$ contributes two terms, $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$ ; and an irreducible quadratic $(ex^2+fx+g)$ , one with no real roots, contributes a term with a linear numerator, $\dfrac{Cx+D}{ex^2+fx+g}$ , never a bare constant.

Figure

A denominator factor on the left mapped by an arrow to its matching partial fraction template on the right, shown for a linear factor, a repeated linear factor, and an irreducible quadratic factor

Finding the constants. Clear denominators by multiplying both sides by $Q(x)$ , then find $A, B, \dots$ either by substituting convenient $x$ -values (especially the roots of linear factors, which make every other term vanish) or by comparing coefficients of matching powers of $x$ . When an irreducible quadratic is present there is no real root to substitute, so coefficient comparison is the reliable route. Many problems combine both strategies: substitute at the roots of the linear factors first to knock out those constants quickly, then compare whatever coefficients remain to pin down the rest — this mixed approach is exactly what the competitive-stretch example below needs, since it has one linear factor and one irreducible quadratic factor together. Once split, each term integrates by a standard form: $\dfrac{A}{x-a}\to A\ln|x-a|$ , $\dfrac{A}{(x-a)^2}\to -\dfrac{A}{x-a}$ , and a quadratic-denominator term is split further into a "derivative of denominator" piece (a logarithm) plus a piece matching $\int\dfrac{dx}{ax^2+bx+c}$ (a standard integral covered separately in this chapter). It helps to recognize this technique as one of three main tools for the Integrals chapter — alongside substitution and integration by parts — and to reach for it specifically when the integrand is a clean ratio of polynomials rather than a product involving trigonometric or exponential factors.

Derivation ADecomposing and integrating a rational function with two distinct linear factors

The goal is to show, end to end, how a rational integrand with a factorable denominator of two distinct linear pieces is converted into a sum of two log terms — the most common pattern this technique produces, and the one worth being fastest at, since a correct root-substitution collapses each unknown constant to a one-line calculation.

Step 1. Confirm the fraction is proper (numerator degree less than denominator degree) — here numerator degree $1 <$ denominator degree $2$ , so no long division is needed — then write the general template with unknown constants for each distinct linear factor: $\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$ .

Step 2. Clear denominators by multiplying both sides by $(x-a)(x-b)$ , giving the polynomial identity $P(x) = A(x-b) + B(x-a)$ , which must hold for every value of $x$ .

Step 3. Exploit the roots. Setting $x = a$ makes the $B$ term vanish (its factor $(x-a)$ becomes zero), instantly giving $A$ ; setting $x = b$ isolates $B$ the same way. This substitution trick is what makes distinct-linear-factor cases the fastest to solve.

Step 4. Integrate each term using $\int \dfrac{k}{x-a}\,dx = k\ln|x-a| + C$ , and add the results, giving a sum of natural-log terms as the final antiderivative. This same four-step skeleton — template, clear, substitute roots, integrate log terms — generalizes directly: with three distinct linear factors you would simply carry three constants and three roots through the identical logic instead of two.

Worked example

Evaluate $\displaystyle\int \frac{3x+5}{(x-1)(x+2)}\,dx$ .

Step 1. The denominator has two distinct linear factors, so write $\dfrac{3x+5}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$ .

Step 2. Clear denominators: $3x+5 = A(x+2) + B(x-1)$ .

Step 3. Set $x=1$ : $3(1)+5 = A(3) \Rightarrow 8 = 3A \Rightarrow A = \dfrac83$ . Set $x=-2$ : $3(-2)+5 = B(-3) \Rightarrow -1 = -3B \Rightarrow B = \dfrac13$ .

Step 4. Integrate term by term:

\int \frac{3x+5}{(x-1)(x+2)}\,dx = \frac83\ln|x-1| + \frac13\ln|x+2| + C

Answer: $\dfrac83\ln|x-1| + \dfrac13\ln|x+2| + C$ .

Competitive stretch

For a mixed case with a linear and an irreducible quadratic factor, $\displaystyle\int \frac{2x^2+x+1}{x(x^2+1)}\,dx$ needs the template $\dfrac{A}{x} + \dfrac{Bx+C}{x^2+1}$ . Clearing denominators gives $2x^2+x+1 = A(x^2+1) + Bx^2 + Cx$ . Setting $x=0$ gives $A=1$ ; comparing $x^2$ coefficients gives $2 = A+B \Rightarrow B=1$ ; comparing $x$ coefficients gives $C=1$ . Splitting $\dfrac{x+1}{x^2+1}$ into $\dfrac{x}{x^2+1}+\dfrac{1}{x^2+1}$ and integrating gives $\ln|x| + \tfrac12\ln(x^2+1) + \tan^{-1}x + C$ .

Common traps

Applying partial fractions directly to an improper fraction (numerator degree greater than or equal to denominator degree) without first doing polynomial long division.
Using the wrong template for a repeated linear factor — $(x-a)^2$ needs two terms, $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$ , not just $\dfrac{A}{(x-a)^2}$ .
Using a constant numerator $\dfrac{A}{ax^2+bx+c}$ for an irreducible quadratic factor instead of the correct linear numerator $\dfrac{Ax+B}{ax^2+bx+c}$ .
Forgetting to split the numerator of a quadratic-denominator term into a derivative-of-denominator part plus a constant before integrating — it cannot be one clean logarithm unless the numerator already matches the denominator's derivative.

Ask the tutor: unsure which partial-fraction template fits your denominator, or want a walkthrough of solving for the unknown constants on your own problem?

Why does substituting the root of a linear factor work so well to find the constants?

When you clear denominators, you get a polynomial identity like $P(x) = A(x-b) + B(x-a)$ that must hold for every $x$ , including $x=a$ and $x=b$ themselves — it is an identity, not just an equation at one point, which is exactly why plugging in any value is legitimate. Plugging in $x=a$ makes the factor $(x-a)$ multiplying $B$ vanish, leaving only the $A$ term — so you solve for $A$ in one line instead of a system of equations. It only works cleanly at the roots of the linear factors; for an irreducible quadratic there is no real root, so you switch to comparing coefficients instead. As a habit, always plug your final constants back into the cleared equation and check it holds for one extra value of $x$ — this catches sign slips before they cost you the whole integral.

How do I know whether to use substitution or coefficient comparison?

Use substitution at every real root of a linear factor first — it is faster and gives constants directly with almost no algebra. If any factor is a repeated linear factor or an irreducible quadratic, you will run out of convenient roots, so compare coefficients of matching powers of $x$ on both sides of the cleared equation for whatever constants remain. Most exam problems combine both: substitute at the linear roots, then compare one or two leftover coefficients — exactly the pattern in the competitive-stretch example, where $x=0$ gives $A$ instantly and the $x^2$ and $x$ coefficients then hand over $B$ and $C$ .

What if the denominator does not factor nicely over the rationals?

The CBSE syllabus restricts problems to denominators that factor into the three standard patterns covered here — distinct linear factors, repeated linear factors, and irreducible quadratics with integer or simple rational coefficients. If a denominator does not factor this way, the problem is outside the syllabus scope; recheck your factoring rather than assuming a harder case is intended. In practice, exam denominators are deliberately built from small integer roots or a recognizable perfect-square pattern, so a quick trial of small integer values (or a fast quadratic-formula check for real roots) is usually enough to reveal the correct factorization before you commit to a template.

Integration by partial fractions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads