Integration by partial fractions

Some integrals refuse every substitution you throw at them — until you notice the integrand is just one polynomial sitting on top of another. A rational function like $\dfrac{3x+5}{(x-1)(x+2)}$ cannot be integrated directly, but it can be split into $\dfrac{A}{x-1}+\dfrac{B}{x+2}$ , and each of those pieces is a standard log integral you already know. That splitting trick is integration by partial fractions, and it is the workhorse technique behind the Integrals (Indefinite) section of the ISC Class 12 Calculus unit. Once you can factor a denominator and match the right template to each factor — linear, repeated linear, or irreducible quadratic — almost any rational integrand becomes routine bookkeeping rather than a puzzle.

The core insight

Any proper rational function decomposes into simple fractions matched to its denominator's factors, and each simple fraction integrates by a standard log or $\tan^{-1}$ form.

The core idea

Partial fractions works because every real polynomial factors into linear pieces and irreducible quadratic pieces, and a proper rational function's decomposition is uniquely determined once you match the correct type of numerator to each factor type.

Distinct linear factor $(x-a)$ contributes a term $\dfrac{A}{x-a}$ .
Repeated linear factor $(x-a)^k$ contributes $k$ separate terms, one for each power up to $k$ — never just one term for the whole repeated factor.
Irreducible quadratic factor $(x^2+bx+c)$ , which cannot be factored further over the reals, needs a linear numerator $\dfrac{Ax+B}{x^2+bx+c}$ , not a constant.

For a proper fraction with two distinct linear factors, the setup is

\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}

where: $P(x)$ is a polynomial of degree lower than the denominator, and $A$ , $B$ are constants found by clearing denominators and substituting $x=a$ and $x=b$ in turn. If the fraction is improper — numerator degree greater than or equal to denominator degree — you must divide first:

\frac{P(x)}{Q(x)} = (\text{quotient}) + \frac{R(x)}{Q(x)}

where: the quotient is an ordinary polynomial from long division, and $R(x)$ , the remainder, has degree strictly less than $Q(x)$ , so $\dfrac{R(x)}{Q(x)}$ is now proper and can be decomposed normally. Skipping this check is the single most common reason a decomposition attempt fails outright. The constants in any decomposition can be found two ways, and most problems need a blend of both. Root substitution works whenever a linear factor's root makes the identity collapse to a single unknown — always use it first, for every distinct or repeated linear factor, since it is faster and less error-prone than algebra. Coefficient comparison is the fallback for whatever root substitution cannot reach directly: the coefficients belonging to an irreducible quadratic factor, or the lower-power constants of a repeated linear factor once its highest power has already been pinned down by substitution. Mixing the two — substitute first for the easy roots, then compare coefficients for what remains — is the standard exam workflow.

Figure

Three denominator factor types matched to their partial fraction templates: linear gives one term, repeated linear gives one term per power, irreducible quadratic gives a linear-over-quadratic term

Derivation ASolving for the constants in a linear-factor decomposition

Deriving the working rule for a proper fraction over two distinct linear factors matters because it is the pattern every other case — repeated linear factors, irreducible quadratic factors, or a mix of both in the same denominator — extends directly, so mastering this one case unlocks the whole technique rather than being a special case you memorise in isolation and forget once the denominator changes shape.

Step 1. Start from the assumed decomposition $\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$ , where $P(x)$ has degree at most 1 (a proper fraction over a quadratic denominator).

Step 2. Multiply both sides by the full denominator $(x-a)(x-b)$ to clear all fractions:

P(x) = A(x-b) + B(x-a)

This is now a polynomial identity, true for every value of $x$ , not just the ones in the domain of the original fraction.

Step 3. Because the identity holds for every $x$ , choose the two values that make one bracket vanish. Put $x=a$ : the $B(x-a)$ term disappears, leaving $P(a) = A(a-b)$ , so $A = \dfrac{P(a)}{a-b}$ . Put $x=b$ : the $A(x-b)$ term disappears, leaving $P(b) = B(b-a)$ , so $B = \dfrac{P(b)}{b-a}$ .

Step 4. With $A$ and $B$ known, each simple fraction integrates by the standard log form $\displaystyle\int \frac{dx}{x-a} = \log|x-a| + C$ , so the full result is

\int \frac{P(x)}{(x-a)(x-b)}\,dx = A\log|x-a| + B\log|x-b| + C

where: $A$ and $B$ are the constants from Step 3, and the absolute values are needed because $\log$ is undefined for negative arguments while $x-a$ or $x-b$ can be negative over parts of the domain. The same root-substitution trick extends to repeated linear factors (substitute the repeated root to get the highest-power constant directly, then compare coefficients for the rest) and to irreducible quadratics, where after clearing denominators you instead compare coefficients of matching powers of $x$ since there is no real root to substitute.

Worked example

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

Step 1. Write $\dfrac{3x+5}{(x-1)(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{x+2}$ — two distinct linear factors, so one constant each.

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

Step 3. Put $x=1$ : $8 = 3A \Rightarrow A = \dfrac{8}{3}$ . Put $x=-2$ : $-1 = -3B \Rightarrow B = \dfrac{1}{3}$ .

Step 4. Integrate each term using the standard log form:

\int\frac{3x+5}{(x-1)(x+2)}\,dx = \frac{8}{3}\log|x-1| + \frac{1}{3}\log|x+2| + C

Final answer: $\dfrac{8}{3}\log|x-1| + \dfrac{1}{3}\log|x+2| + C$ .

Competitive stretch

For a repeated linear factor, such as $\dfrac{x^2+1}{(x-1)^2(x+1)}$ , the template needs three terms: $\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+1}$ . Substituting $x=1$ and $x=-1$ gives $B=1$ and $C=\tfrac12$ directly (the repeated root kills the other terms cleanly at its highest power), and comparing the $x^2$ coefficient then gives $A=\tfrac12$ . The integral comes out to $\tfrac12\log|x-1| - \dfrac{1}{x-1} + \tfrac12\log|x+1| + C$ , mixing a log term with a power term — a signature of repeated factors.

Common traps

Skipping the improper-fraction check. If numerator degree is greater than or equal to denominator degree, divide first — decomposing directly gives an unsolvable system for the constants.
Using one constant for a repeated factor instead of one term per power, e.g. writing only $\dfrac{A}{(x-1)^2}$ instead of $\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}$ .
Mismatching numerator type to factor type — a constant numerator over an irreducible quadratic, or a linear numerator over a simple linear factor, both break the decomposition.
Sign slips when substituting negative roots — redo the arithmetic explicitly rather than trusting a mental shortcut, especially for roots like $x=-2$ .

Ask the tutor: Not sure which template fits a mixed denominator like $(x-1)(x^2+4)$ ? Ask the AI tutor to walk through setting up each term with you.

Why can't I just use a constant numerator for every factor, linear or quadratic?

Because the numerator's degree must always be one less than the factor it sits over. A linear factor $(x-a)$ has degree 1, so its numerator is a constant (degree 0). An irreducible quadratic $(x^2+bx+c)$ has degree 2, so its numerator must be degree 1, i.e. $Ax+B$ . Using a constant there leaves the system of equations for the unknowns unsolvable, because you have too few constants to match every power of $x$ once you clear denominators.

How do I know if a quadratic factor is irreducible instead of splitting into two linear factors?

Check its discriminant. For $x^2+bx+c$ , compute $b^2-4c$ . If this is negative, the quadratic has no real roots and cannot be factored further over the reals — it stays as one irreducible quadratic factor with an $Ax+B$ numerator. If the discriminant is zero or positive, factor it into linear pieces first and use the ordinary linear-factor templates instead, since that decomposition is simpler to integrate.

Why does the root-substitution trick not work for repeated factors past the highest power?

Substituting $x=a$ into the cleared identity for $(x-a)^k$ kills every term except the one attached to the highest power, $(x-a)^k$ , giving that one constant directly. But the lower-power constants are still multiplied by other factors when $x=a$ is substituted, so they do not vanish to zero the same way — you need a second method, usually comparing coefficients of matching powers of $x$ , to pin those down.

Integration by partial fractions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads