Splitting a complicated fraction into simple pieces sounds like an exam-only ritual, yet it is exactly how engineers decode the behaviour of circuits, autopilots and camera stabilisers — and how a computer integrates any rational function you hand it. The move behind all of that is surprisingly homely.

You have spent years adding fractions: put them over a common denominator, combine, tidy up. Partial fractions is that skill played backwards — un-adding. Watch what happens when we add two utterly harmless fractions:

So \dfrac{1}{x^2-1} is secretly two simple fractions in a trench coat. Why does a calculus course care? Because as written, \displaystyle\int \frac{dx}{x^2-1} looks hopeless — it is not a power, not a standard form, and no obvious substitution cracks it. But each piece integrates to a logarithm in one line:

That is the whole method in one sentence: factor the denominator, split the fraction into its simple parts, integrate each part in one line. A single algebraic move unlocks an entire family of integrals — every rational function \dfrac{P(x)}{Q(x)} whose denominator factors becomes a sum of pieces you already know how to integrate. The calculus here is trivial; the craft is in the un-adding. This page is about that craft.

Worked example 1: \displaystyle\int \frac{dx}{(x-1)(x+2)}

Step 1 — write the decomposition with unknowns. Two distinct linear factors, so two simple fractions:

Step 2 — clear the denominators. Multiply both sides by (x-1)(x+2):

This is an identity — it must hold for every x, not just some solution. That single fact gives us two different ways to hunt down A and B. Learn both: one is fast, the other never lets you down.

Route 1 — substitute the roots (the cover-up idea)

Since the identity holds for every x, pick values that make terms vanish. Setting x = 1 kills the B term (x - 1 = 0):

Setting x = -2 kills the A term:

Route 2 — equate coefficients

Alternatively, expand the right side and gather powers of x:

The left side is 0 \cdot x + 1. Two polynomials are identical only if their coefficients match, power by power:

Solve the pair: adding gives 3A = 1, so A = \tfrac13 and B = -\tfrac13 — the same answer, as it must be. Substituting roots is quicker when the factors are distinct and linear; equating coefficients works for every shape of decomposition, and doubles as a check on your arithmetic.

Finish — integrate each piece

and each piece is a one-line logarithm (via u = x - 1 and u = x + 2):

The shapes of decomposition

Distinct linear factors are the easy case; two more shapes round out the method. The guiding principle: each factor of the denominator contributes its own template of unknowns, and the total number of unknowns always equals the degree of the denominator.

The linear-factor pieces integrate to logarithms; the higher powers \dfrac{A}{(x-r)^k} integrate by the power rule to \dfrac{-A}{(k-1)(x-r)^{k-1}}; and the quadratic pieces split into a logarithm plus an \arctan. Every proper rational function, without exception, dissolves into these three kinds of one-line integrals — which is why partial fractions is the closing move of the integration toolkit: it guarantees that every rational function can be integrated in elementary terms.

Worked example 2: a repeated factor — \displaystyle\int \frac{x+2}{(x-1)^2}\,dx

The denominator is (x-1)^2: one linear factor, squared. The template demands a term for each power:

Multiply through by (x-1)^2:

Substituting the root x = 1 gives 3 = B — note that the root only reaches the highest power's constant; that is why cover-up alone cannot finish a repeated-factor problem. For A, equate the x-coefficients: 1 = A. So

Now integrate: the first piece is a log, and the second is just the power rule in disguise (3(x-1)^{-2} integrates to -3(x-1)^{-1}):

A quick sanity check never hurts: differentiate the answer — \frac{1}{x-1} + \frac{3}{(x-1)^2} — and you are looking at the decomposition again. The circle closes.

Worked example 3: an improper fraction — \displaystyle\int \frac{x^2}{(x-1)(x+2)}\,dx

Degree check first: the top has degree 2, the bottom has degree 2 — top \ge bottom, so this fraction is improper and we must divide before we may decompose. Expanding the denominator, (x-1)(x+2) = x^2 + x - 2, and dividing:

The polynomial part 1 integrates trivially; the remainder is now proper (degree 1 over degree 2), so the templates apply. Cover-up on \dfrac{2-x}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}:

(Check by coefficients: A + B = \tfrac13 - \tfrac43 = -1 matches the -x, and 2A - B = \tfrac23 + \tfrac43 = 2 matches the constant. It works.) Assemble and integrate term by term:

Three moves — divide, decompose, integrate — and a fraction that offered no way in has become a polynomial plus two logarithms.

See it: a rational function and its antiderivative

The faint curve is the integrand \dfrac{1}{(x-1)(x-s)}, blowing up at its two vertical asymptotes x = 1 and x = s. The bold curve is the antiderivative that partial fractions hands us, \dfrac{1}{1-s}\big(\ln|x-1| - \ln|x-s|\big) — cover-up gives A = \tfrac{1}{1-s} and B = -\tfrac{1}{1-s}, so the two log coefficients are always equal and opposite.

Slide the right root s and watch the pair respond. Two things are worth spotting. First, the antiderivative has its own vertical asymptotes at the very same places — where the integrand blows up, its accumulated area blows up too, so \ln|x-r| plunges to -\infty at each root. Second, drag s down towards 1: the coefficient \tfrac{1}{1-s} grows without bound as the two roots close in on each other. The distinct-roots formula is tearing itself apart — a preview of why a repeated root demands the different template \tfrac{A}{x-r} + \tfrac{B}{(x-r)^2} you met in worked example 2.

See it explained