Partial fractions — linear, repeated, irreducible quadratic factors: Visual walkthrough

Step 0 — What is a "rational function", and what are we allowed to touch?

We will trace one running example the whole way: $P (x) 3 x + 1 / Q (x) (x^{2} - x - 2) .$

$P (x) = 3 x + 1$ — the top, degree $1$ .
$Q (x) = x^{2} - x - 2$ — the bottom, degree $2$ .

Because $de g P = 1 < 2 = de g Q$ , this fraction is proper — the top is genuinely smaller than the bottom. Why we check this first: only proper fractions can be built purely from simple pieces; an oversized top would need a leftover polynomial (that is Polynomial long division, handled separately). Every step below assumes proper.

Step 1 — See the denominator "un-multiply" into factors

WHAT. We split the bottom into a product of the simplest possible factors: $x^{2} - x - 2 = (x - 2) (x + 1) .$ Here $x - 2$ and $x + 1$ are linear factors — each is just " $x$ minus a number". The numbers $2$ and $- 1$ are the roots: the $x$ -values that make the bottom zero (this is Factoring polynomials over the reals).

WHY. The denominator is a product. A product of two things is only zero when one of them is zero. So the whole fraction blows up ("explodes to infinity") at exactly two places: $x = 2$ and $x = - 1$ . Those two explosion points are the fingerprints we will chase.

PICTURE. Below, the blue curve is $y = \frac{3 x + 1}{( x - 2 ) ( x + 1 )}$ . Notice the two vertical dashed walls — one yellow at $x = - 1$ , one red at $x = 2$ — where the curve shoots off to infinity. Each wall is caused by one factor turning zero.

Figure — Partial fractions — linear, repeated, irreducible quadratic factors

Step 2 — Guess the shape: one simple fraction per explosion

WHAT. We claim the fraction is a sum, one term per factor: $\frac{3 x + 1}{( x - 2 ) ( x + 1 )} = \frac{A}{x - 2} + \frac{B}{x + 1} .$

$\frac{A}{x - 2}$ — a piece that explodes only at $x = 2$ (its own wall).
$\frac{B}{x + 1}$ — a piece that explodes only at $x = - 1$ .
$A$ and $B$ — two unknown numbers ("how strong is each explosion") we must still find.

WHY this exact shape. Adding fractions is reversible. If someone added $\frac{A}{x - 2} + \frac{B}{x + 1}$ over a common denominator, they would land back on $(x - 2) (x + 1)$ on the bottom. So going backwards, our fraction must have come from pieces with those very denominators. The top of each piece is a lone constant because each denominator is degree $1$ , and a proper piece needs a top of degree $0$ .

PICTURE. The blue curve (the full function) is drawn as the sum of the two coloured simple curves: the red one owns the $x = 2$ wall, the yellow one owns the $x = - 1$ wall. Away from the walls they gently add up to the blue.

Recall

Why is the top of each simple piece just a constant (not $A x + \dots$ )? ::: Each denominator has degree $1$ , so a proper piece needs a numerator of degree $0$ — a lone constant.

Step 3 — The cover-up trick, watched term by term

WHAT. To find $A$ , multiply both sides by the factor $(x - 2)$ : $\frac{3 x + 1}{x + 1} = A + (x - 2) \cdot \frac{B}{x + 1} .$

Look at each piece:

Left: the $(x - 2)$ cancelled the $(x - 2)$ that was in the bottom, leaving $\frac{3 x + 1}{x + 1}$ .
Right, term 1: $\frac{A}{x - 2} \times (x - 2) = A$ — bare, no $x$ left.
Right, term 2: $\frac{B}{x + 1} \times (x - 2) = (x - 2) \frac{B}{x + 1}$ — still carries a live $(x - 2)$ .

WHY. We engineered the situation so that at $x = 2$ the annoying $B$ -term is multiplied by $(x - 2) = 0$ and vanishes. That leaves $A$ alone: $A = \frac{3 x + 1}{x + 1}_{x = 2} = \frac{3 ( 2 ) + 1}{2 + 1} = \frac{7}{3} .$

PICTURE. The animation-frame shows the $B$ -term's contribution collapsing to zero as $x$ slides to $2$ (green curve pinched to the axis), while the surviving quantity $\frac{3 x + 1}{x + 1}$ (blue) is read off cleanly at the yellow dot.

Repeat, covering $(x + 1)$ and setting $x = - 1$ : $B = \frac{3 x + 1}{x - 2}_{x = - 1} = \frac{- 3 + 1}{- 1 - 2} = \frac{- 2}{- 3} = \frac{2}{3} .$

Step 4 — Assemble and sanity-check the split

WHAT. With $A = \frac{7}{3}$ , $B = \frac{2}{3}$ : $\frac{3 x + 1}{( x - 2 ) ( x + 1 )} = \frac{7/3}{x - 2} + \frac{2/3}{x + 1} .$

WHY check. Two unknowns, two conditions — but let us verify by testing a safe point $x = 0$ (far from both walls):

Left: $\frac{1}{( - 2 ) ( 1 )} = - \frac{1}{2}$ .
Right: $\frac{7/3}{- 2} + \frac{2/3}{1} = - \frac{7}{6} + \frac{4}{6} = - \frac{3}{6} = - \frac{1}{2}$ . ✓

PICTURE. Blue (original) and the coloured sum are drawn on the same axes; they lie exactly on top of each other everywhere except the two walls. The green vertical tie-line at $x = 0$ shows both giving $- \frac{1}{2}$ .

Step 5 — Integrate: why each simple wall becomes a logarithm

WHAT. $\int \frac{3 x + 1}{( x - 2 ) ( x + 1 )} d x = \frac{7}{3} ln ∣ x - 2∣ + \frac{2}{3} ln ∣ x + 1∣ + C .$

WHY log? The only function whose derivative is $\frac{1}{x - a}$ is $ln ∣ x - a ∣$ — that is a standard integral. So a fraction that behaves like $\frac{constant}{x - a}$ near its wall must integrate to a scaled log. This is the whole payoff: we traded one un-integrable mess for two instant logs.

PICTURE. Each coloured $\frac{1}{x - a}$ spike (top) sits above its own $ln ∣ x - a ∣$ curve (bottom), with the vertical wall at $x = a$ becoming the point where the log dives to $- \infty$ .

Step 6 — Edge case: what if the bottom doesn't split into linear walls?

WHAT. Suppose the bottom carries an irreducible quadratic like $x^{2} + 1$ — one that has no real roots (its discriminant $0^{2} - 4 \cdot 1 \cdot 1 = - 4 < 0$ ). Then there is no real wall for it: the curve $\frac{1}{x ^{2} + 1}$ never explodes.

WHY it needs a linear top $B x + C$ . With no real root to "cover up", we cannot isolate a single constant. The quadratic denominator has degree $2$ , so a proper piece over it needs a numerator of degree $1$ : $\frac{B x + C}{x ^{2} + 1}$ . The $B x$ part integrates to a log (via Integration by substitution, $u = x^{2} + 1$ ); the leftover constant $C$ integrates to an arctan — the smooth "no-wall" swirl, another standard integral. See Completing the square for the general $x^{2} + b x + c$ .

PICTURE. Contrast panel: left, $\frac{1}{x - 1}$ with its real wall (→ log); right, $\frac{1}{x ^{2} + 1}$ , a smooth hump with no wall (→ arctan). Different geometry ⇒ different integral.

The one-picture summary

One messy blue curve, two explosion-walls → two coloured simple fractions found by cover-up → two logs when integrated. That is the entire machine.

Recall Feynman retelling — the whole walkthrough in plain words

I have one lumpy fraction. First I look at its bottom and factor it — every factor marks a spot where the graph shoots to infinity, a "wall". I guess that my lumpy fraction is really a sum, one baby fraction per wall, each baby exploding at only its own wall. To find how strong a baby is, I use the cover-up trick: I multiply through by that wall's factor so every other baby gets multiplied by something that goes to zero right at the wall — they all disappear, and the survivor hands me its number. I do this for each wall, glue the babies back together, and test one easy point to make sure they add to the original. Finally I integrate: each $\frac{1}{x - a}$ wall is just $ln ∣ x - a ∣$ . If instead the bottom had a piece with no real wall (an irreducible quadratic), that piece keeps a linear top $B x + C$ and integrates to a log plus a smooth arctan swirl instead. Walls → logs, no-walls → arctans. Done.

Active recall

Why does one factor of the denominator correspond to one "wall" on the graph?

The fraction explodes to infinity wherever the bottom is zero; each factor supplies exactly one such zero.

In the cover-up step, why does the other term vanish?

You multiply by

(x - a)

; the other term keeps a live factor

(x - a)

which becomes

0

x = a

For our example, what are

A

and

B

A = \frac{7}{3}

over

(x - 2)

and

B = \frac{2}{3}

over

(x + 1)

Why does each simple piece integrate to a log?

\int \frac{1}{x - a} d x = ln ∣ x - a ∣

is the standard integral matching a lone constant over a linear factor.

Why does an irreducible quadratic keep a

B x + C

top and give an arctan?

It has no real root to cover up; degree-2 bottom needs a degree-1 top, and the leftover constant integrates to arctan (no real wall).