partial fraction

Partial fraction decomposition

Let $\frac{P(x)}{Q(x)}$ be a rational function ($P,Q$ polynomials) with $\mathrm{deg}P<\mathrm{deg}Q$.
Factor the denominator into linear factors: $Q(x)=(x-r_1)(x-r_2)\cdots (x-r_n)$.
The $r_i$ are the roots of $Q$ (factor theorem).
Leading coefficients can be absorbed into the numerator.
Assume the roots are distinct, so that no factor repeats; repeated and complex roots change the shape of the pieces, see below.
Then there exist constants $A_1,\dots ,A_n$ such that

$$\frac{P(x)}{(x-r_1)(x-r_2)\cdots (x-r_n)}=\frac{A_1}{x-r_1}+\frac{A_2}{x-r_2}+\cdots +\frac{A_n}{x-r_n}$$

and exactly one choice of the $A_i$ works.

Adding requires bringing every summand to the common denominator: multiply top and bottom of each summand by the factor it is missing, then add the numerators, e.g.:

$$\frac{1}{x-2}+\frac{1}{x+1}=\frac{1\cdot (x+1)}{(x-2)(x+1)}+\frac{1\cdot (x-2)}{(x+1)(x-2)}=\frac{(x+1)+(x-2)}{(x-2)(x+1)}=\frac{2x-1}{(x-2)(x+1)}$$

In the result, the denominator is the product of all factors, and the numerator is a sum in which each original numerator got multiplied by the other factors. Its degree stays below the denominator’s (each numerator term has $n-1$ of the $n$ factors), hence the $\mathrm{deg}P<\mathrm{deg}Q$ condition in the definition.
The decomposition inverts this computation: it recovers the summands from the combined fraction.

Each summand can be integrated, summed, or bounded on its own; the combined fraction can't.

Computing the decomposition of $\frac{3x+1}{x^2-x-2}$

Factor the denominator: $x^2-x-2=(x-2)(x+1)$, roots $r_1=2$ and $r_2=-1$.
Write the decomposition with unknown constants:

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

Multiply both sides by $(x-2)(x+1)$. On the right, each fraction cancels its own factor and keeps the other:

$$3x+1=A(x+1)+B(x-2)$$

The identity holds for every $x$, so plugging in any value gives a valid equation in $A,B$.
But plug in e.g. $x=0$ and $x=1$, and the equations are $1=A-2B$ and $4=2A-B$: each contains both unknowns, and they must be solved together (isolate $A$ in one, substitute into the other, …).
Plugging in a root instead makes every term except one vanish ($r_i-r_i=0$), so each equation contains a single unknown.

Why plugging in the roots is allowed $x=2,-1$ (division by zero), so after multiplying through, the polynomial equation is so far only known to hold for $x\ne 2,-1$. Two polynomials that agree at infinitely many points are identical: their difference is a polynomial with infinitely many roots, and a nonzero polynomial of degree $n$ has at most $n$ roots (factor theorem). So the equality extends to $x=2,-1$.

The fraction equation above is undefined at

$$x=2:3\cdot 2+1=A(2+1)+B\underset{=0}{\underbrace{(2-2)}}⟹7=3A⟹A=\frac{7}{3}$$ $$x=-1:3\cdot (-1)+1=A\underset{=0}{\underbrace{(-1+1)}}+B(-1-2)⟹-2=-3B⟹B=\frac{2}{3}$$

Result:

$$\frac{3x+1}{x^2-x-2}=\frac{7/3}{x-2}+\frac{2/3}{x+1}$$

Check by adding the right side back together: $\frac{7}{3}(x+1)+\frac{2}{3}(x-2)=\frac{9}{3}x+\frac{3}{3}=3x+1$ ✓

Decompose $\frac{1}{k(k+1)}$ into partial fractions.

$$\frac{1}{k(k+1)}=\frac{A}{k}+\frac{B}{k+1}⟹1=A(k+1)+Bk$$

Plug in $k=0⟹A=1$
Plug in $k=1⟹B=-1$

$$⟹\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$$

Cleanup slop below

TODO, show all 3: Plugging in roots, Comparing coefficients, Rewriting the numerator

Ansatz shapes for general denominators

Over $\mathbb{R}$, every polynomial factors into linear terms and quadratics without real roots (fundamental theorem of algebra). Each factor of $Q$ gets one piece of the ansatz:

factor of $Q$	piece	unknowns
$(x-r)$	$\frac{A}{x-r}$	$1$
$(x-r{)}^m$	$\frac{A_1}{x-r}+\frac{A_2}{(x-r{)}^2}+\cdots +\frac{A_m}{(x-r{)}^m}$	$m$
$x^2+px+q$, no real root	$\frac{Ax+B}{x^2+px+q}$	$2$

A repeated factor $(x-r{)}^m$ needs all $m$ powers because its share of the numerator can be any polynomial of degree $<m$, and a single constant can’t express that. $\frac{x}{(x-1{)}^2}$ shows the failure: no constant $A$ makes $\frac{A}{(x-1{)}^2}$ equal to it. Rewriting the numerator around the root, $x=(x-1)+1$, produces both powers:

$$\frac{x}{(x-1{)}^2}=\frac{(x-1)+1}{(x-1{)}^2}=\frac{1}{x-1}+\frac{1}{(x-1{)}^2}$$

If $\mathrm{deg}P\ge \mathrm{deg}Q$

The fraction has a polynomial “whole part”, like $\frac{7}{3}=2+\frac{1}{3}$ for integers.
Split it off first by polynomial long division; the decomposition applies to the remainder fraction.

Used wherever a product sits in a denominator and the pieces are tractable one at a time:
Integration … each piece integrates to a logarithm or arctan.
telescoping sums … $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ collapses the partial sums of $\sum \frac{1}{k(k+1)}$ to $1-\frac{1}{n+1}$.

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Old slop below

Partial fractions decompose rational functions (fractions with polynomials) into simpler fractions that are easier to integrate.

When to Use

For integrals of the form $\int \frac{P(x)}{Q(x)}dx$ where:

• $P(x)$ and $Q(x)$ are polynomials
• Degree of $P(x)$ < degree of $Q(x)$ (otherwise, do polynomial long division first)

The Method

1. Factor the denominator $Q(x)$ completely
2. Write the partial fraction form based on factor types
3. Find the constants using algebra
4. Integrate each simple fraction

Forms for Different Factor Types

Linear Factors: $(ax+b)$

$$\frac{A}{ax+b}$$

Repeated Linear Factors: $(ax+b{)}^n$

$$\frac{A_1}{ax+b}+\frac{A_2}{(ax+b{)}^2}+...+\frac{A_n}{(ax+b{)}^n}$$

Irreducible Quadratic: $(a{x}^2+bx+c)$

$$\frac{Ax+B}{a{x}^2+bx+c}$$

Example: $\int \frac{3x+1}{x^2-x-2}dx$

Step 1: Factor denominator

$$x^2-x-2=(x-2)(x+1)$$

Step 2: Set up partial fractions

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

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

$$3x+1=A(x+1)+B(x-2)$$

Substitute convenient values:

• Let $x=2$: $7=3A$ → $A=\frac{7}{3}$
• Let $x=-1$: $-2=-3B$ → $B=\frac{2}{3}$

Step 4: Integrate

$$\int \left(\frac{7/3}{x-2}+\frac{2/3}{x+1}\right)dx=\frac{7}{3}\ln |x-2|+\frac{2}{3}\ln |x+1|+C$$

Basic arithmetic example

$$\frac{3}{x}+\frac{1}{y}=\frac{3y}{xy}+\frac{x}{xy}=\frac{3y+x}{xy}$$

See also: ^e2cb34